整理各類模板（準備隨時棄坑）

線段樹1（真題）

#include<iostream>
#include<cstring>
#include<cmath>
#include<cstdio> 
using namespace std;
const int N=1e5+100;
long long tr[N*4],flag[N*4],ans;
void built(int i,int l,int r,int x,long long val)
{
    if(l==r)
    {
        tr[i]+=val; 
        return;
    }
    int mid=(l+r)/2;
    if(x<=mid) built(i*2,l,mid,x,val); 
    if(x>mid) built(i*2+1,mid+1,r,x,val); 
    tr[i]=tr[i*2]+tr[i*2+1]; 
    return;
}
void pushdown(int i,int l,int r) 
{
    int mid=(l+r)/2;
    tr[i*2]+=flag[i]*(mid-l+1);
    tr[i*2+1]+=flag[i]*(r-mid); 
    flag[i*2]+=flag[i];
    flag[i*2+1]+=flag[i];
    flag[i]=0; 
    return;
}
void update(int i,int l,int r,long long x,long long y,long long k) 
{
    if(l>=x&&r<=y)
    {
        tr[i]+=k*(r-l+1);
        flag[i]+=k; 
        return;
    }
    if(flag[i]>0) pushdown(i,l,r);
    int mid=(l+r)/2;
    if(mid>=x) update(i*2,l,mid,x,y,k); 
    if(mid<y) update(i*2+1,mid+1,r,x,y,k);
    tr[i]=tr[i*2]+tr[i*2+1];
    return;
}
void query(int i,int l,int r,long long x,long long y)
{
    if(l>=x&&r<=y)
    {
        ans+=tr[i]; 
        return;
    }
    if(flag[i]>0) pushdown(i,l,r); 
    int mid=(l+r)/2;
    if(mid>=x) query(i*2,l,mid,x,y); 
    if(mid<y) query(i*2+1,mid+1,r,x,y); 
}
int main()
{
    int n,m;
    cin>>n>>m;
    for(int i=1;i<=n;i++)
    {
        long long x;
        cin>>x;
        built(1,1,n,i,x);
    }
    while(m--)
    {
        int op;
        cin>>op;
        if(op==1)
        {
            long long x,y,k;
            cin>>x>>y>>k;
            update(1,1,n,x,y,k);
        }
        if(op==2)
        {
            long long x,y;
            cin>>x>>y;
            ans=0;
            query(1,1,n,x,y);
            cout<<ans<<endl;
        }
    }
    return 0;
}

線段樹2（真題）

#include<bits/stdc++.h>
using namespace std;
const int N = 1e5+100;
int n,m;
int a,b,c;
long long sum[N*4],fa[N*4],fm[N*4],w,p,ans;
bool mu[N*4];
int mid[N*4];
void build(int i,int l,int r)
{
    fm[i]=1;
    if(l==r)
    {
        cin>>sum[i];
        return;
    }
    mid[i]=(l+r)/2;
    build(i*2,l,mid[i]);
    build(i*2+1,mid[i]+1,r);
    sum[i]=(sum[i*2]+sum[i*2+1])%p;
}
void down(int i,int l,int mid,int r)
{
    if(mu[i])
    {
        sum[i*2]=sum[i*2]*fm[i]%p;
        sum[i*2+1]=sum[i*2+1]*fm[i]%p;
        fa[i*2]=fa[i*2]*fm[i]%p;
        fa[i*2+1]=fa[i*2+1]*fm[i]%p;
        fm[i*2]=fm[i*2]*fm[i]%p;
        fm[i*2+1]=fm[i*2+1]*fm[i]%p;
        mu[i*2]=mu[i*2+1]=true;
        fm[i]=1;
        mu[i]=false;    
    }
    if(fa[i])
    {
        sum[i*2]=(sum[i*2]+fa[i]*(mid-l+1)%p)%p;
        sum[i*2+1]=(sum[i*2+1]+fa[i]*(r-mid)%p)%p;
        fa[i*2]=(fa[i*2]+fa[i])%p;
        fa[i*2+1]=(fa[i*2+1]+fa[i])%p;
        fa[i]=0;
    }
}
void solve(int i,int l,int r)
{
    if(l>=b && r<=c)
    {
        if(a==1)
        {
            sum[i]=sum[i]*w%p;
            fa[i]=fa[i]*w%p;
            fm[i]=fm[i]*w%p;
            mu[i]=true;
        }
        else
        if(a==2)
        {
            sum[i]=(sum[i]+w*(r-l+1)%p)%p;
            fa[i]+=w;
        }
        else
        ans=(ans+sum[i])%p;
        return;
    }
    if(mu[i] || fa[i]) down(i,l,mid[i],r);
    if(b<=mid[i]) solve(i*2,l,mid[i]);
    if(c>mid[i]) solve(i*2+1,mid[i]+1,r);
    sum[i]=(sum[i*2]+sum[i*2+1])%p;
}
int main()
{
    cin>>n>>m>>p;
    build(1,1,n);
    while(m--)
    {
        cin>>a;
        if(a==1 || a==2)
        {
            cin>>b>>c>>w;
            solve(1,1,n);
        }
        else
        {
            ans=0;
            cin>>b>>c;
            solve(1,1,n);
            cout<<ans<<endl;
        }
    }
    return 0;
}

ST表（真題）

#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
inline int read()
{
    char c=getchar();
    int x=0,f=1;
    while(c<'0'||c>'9')
    {
        if(c=='-')f=-1;
        c=getchar();
    }
    while(c>='0'&&c<='9')
    {
        x=x*10+c-'0';
        c=getchar();
    }
    return x*f;
}
int maxx[1000010][21];
int query(int l,int r)
{
    int k=log2(r-l+1);
    return max(maxx[l][k],maxx[r-(1<<k)+1][k]);
}
int main()
{
    int n=read(),m=read();
    for(int i=1; i<=n; i++)
        maxx[i][0]=read();
    for(int j=1; j<=21; j++)
        for(int i=1; i+(1<<j)-1<=n; i++)
            maxx[i][j]=max(maxx[i][j-1],maxx[i+(1<<(j-1))][j-1]);
    for(int i=1; i<=m; i++)
    {
        int l=read(),r=read();
        printf("%d\n",query(l,r));
    }
    return 0;
}

並查集（真題）

#include<bits/stdc++.h>
using namespace std;
int n,m;
int x,y,z;
int a[11000];
int find(int x)
{
    if(x==a[x])
        return x;
    else
    return a[x]=find(a[x]);
}
int main()
{
    cin>>n>>m;
    while(n--)
        a[n]=n;
    while(m--)
    {
        cin>>x>>y>>z;
        if(x&2)
        {
            if(find(y)==find(z))
                cout<<"Y"<<endl;
            else
                cout<<"N"<<endl;
        }
        else
        if(find(y)!=find(z))
            a[find(y)]=find(z);
    }
    return 0;
}

乘法逆元（真題）

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#define int long long
using namespace std;
int a[3000010];
int n,p;
main()
{
    cin>>n>>p;
    a[1]=1;
    cout<<1<<"\n";
    for(int i=2;i<=n;i++)
    {
        a[i]=(p-p/i)*a[p%i]%p;
        cout<<a[i]<<"\n";
    }
    return 0;
}

單源最短路（真題）

#include <bits/stdc++.h>
#define re register
using namespace std;
inline int read() {
    int X=0,w=1; char c=getchar();
    while (c<'0'||c>'9') { if (c=='-') w=-1; c=getchar(); }
    while (c>='0'&&c<='9') X=(X<<3)+(X<<1)+c-'0',c=getchar();
    return X*w;
}
struct Edge { int v,w,nxt; };
Edge e[500010];
int head[100010],cnt=0;
inline void addEdge(int u,int v,int w) {
    e[++cnt].v=v;
    e[cnt].w=w;
    e[cnt].nxt=head[u];
    head[u]=cnt;
}
int n,m,s;
int dis[100010];
struct node { //堆節點
    int u,d;
    bool operator <(const node& rhs) const {
        return d>rhs.d;
    }
};
inline void Dijkstra() {
    for (re int i=1;i<=n;i++) dis[i]=2147483647;
    dis[s]=0;
    priority_queue<node> Q; //堆
    Q.push((node){s,0});
    while (!Q.empty()) {
        node fr=Q.top(); Q.pop();
        int u=fr.u,d=fr.d;
        if (d!=dis[u]) continue;
        for (re int i=head[u];i;i=e[i].nxt) {
            int v=e[i].v,w=e[i].w;
            if (dis[u]+w<dis[v]) {
                dis[v]=dis[u]+w;
                Q.push((node){v,dis[v]});
            }
        }
    }
}
int main() {
    n=read(),m=read(),s=read();
    for (re int i=1;i<=m;i++) {
        int X=read(),Y=read(),Z=read();
        addEdge(X,Y,Z);
    }
    Dijkstra();
    for (re int i=1;i<=n;i++) printf("%d ",dis[i]);
    return 0;
}

堆(真題)

#include<bits/stdc++.h>
using namespace std;
priority_queue< int,vector<int>,greater<int> >a;
int n,m;
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&m);
        if(m==1)
        {
            scanf("%d",&m);
            a.push(m);
        }
        else
        if(m==2)
        {
            cout<<a.top()<<endl;
        }
        else
        a.pop();
    }
    return 0;
}

二分圖匹配（真題）

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
using namespace std;
int map[1010][1010];
int pl[1000010];
bool mark[1000010];
int n,m,e,tot;
inline int read()
{
    int x=0,f=1;
    char ch=getchar();
    for(; !isdigit(ch); ch=getchar())
        if(ch=='-')
            f=-1;
    for(; isdigit(ch); ch=getchar())
        x=(x<<1)+(x<<3)+ch-'0';
    return x*f;
}
bool find(int x)
{
    for(int i=1; i<=m; i++)
    {
        if(map[x][i]==1&&mark[i]==0)
        {
            mark[i]=1;
            if(pl[i]==0||find(pl[i])==1)
            {
                pl[i]=x;
                return 1;
            }
        }
    }
    return 0;
}
int main()
{
    n=read(),m=read(),e=read();
    for(int i=1; i<=e; i++)
    {
        int x=read(),y=read();
        if(x<=n&&y<=m)
            map[x][y]=1;
    }
    for(int i=1; i<=n; i++)
    {
        memset(mark,0,sizeof(mark));
        if(find(i)==1)
            tot++;
    }
    cout<<tot;
    return 0;
}

負環（真題）

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
using namespace std;

typedef long long LL;

inline int read()
{
    char c=getchar();int num=0,f=1;
    for(;!isdigit(c);c=getchar())
        f=c=='-'?-1:f;
    for(;isdigit(c);c=getchar())
        num=num*10+c-'0';
    return num*f;
}

const int N=2e3+5;
const int M=3e3+5;

int T,n,m;

int head[N],num_edge;
struct Edge
{
    int v,w,nxt;
}edge[M<<1];

inline void add_edge(int u,int v,int w)
{
    edge[++num_edge].v=v;
    edge[num_edge].w=w;
    edge[num_edge].nxt=head[u];
    head[u]=num_edge;
}

int dis[N],cnt[N],vis[N],tim;
queue<int> que;
bool spfa()
{
    ++tim;
    memset(dis,0x3f,sizeof(dis));
    memset(cnt,0,sizeof(cnt));
    while(!que.empty()) que.pop();
    int u;dis[1]=0,cnt[1]=1;que.push(1);
    while(!que.empty())
    {
        u=que.front(),que.pop();
        vis[u]=0;
        for(int i=head[u],v;i;i=edge[i].nxt)
        {
            v=edge[i].v;
            if(dis[v]>dis[u]+edge[i].w)
            {
                dis[v]=dis[u]+edge[i].w;
                ++cnt[v]/*=cnt[u]+1*/;
                if(cnt[v]>=n) return 1;
                if(vis[v]!=tim)
                {
                    vis[v]=tim;
                    que.push(v);
                }
            }
        }
    }
    return 0;
}

int main()
{
    T=read();
    while(T--)
    {
        memset(head,0,sizeof(head));num_edge=0;
        n=read(),m=read();
        for(int i=1,a,b,c;i<=m;++i)
        {
            a=read(),b=read(),c=read();
            add_edge(a,b,c);
            if(c>=0) add_edge(b,a,c);
        }
        if(spfa()) puts("YE5");
        else puts("N0");
    }
    return 0;
}

矩陣快速冪（真題）

#include<iostream>
#include<cstring>
#define mod 1000000007
#define ll long long
using namespace std;
struct Mat{
    ll m[101][101];
};
Mat a,e;
ll n,p;
Mat Mul(Mat x,Mat y) 
{
    Mat c;
    for(int i=1;i<=n;i++)
      for(int j=1;j<=n;j++)
        c.m[i][j]=0;
    for(int i=1;i<=n;i++)
      for(int j=1;j<=n;j++)
        for(int k=1;k<=n;k++)
        {
            c.m[i][j]=c.m[i][j]%mod+x.m[i][k]*y.m[k][j]%mod;
          }
    return c; 
}
Mat pow(Mat x,ll y)
{
    Mat ans=e;
    while(y)
    {
        if(y&1)
         ans=Mul(ans,x);  
        x=Mul(x,x);
        y>>=1;
    }
    return ans;
}

int main()
{
    cin>>n>>p;
    for(int i=1;i<=n;i++)
      for(int j=1;j<=n;j++)
        cin>>a.m[i][j];
    for(int i=1;i<=n;i++)
        e.m[i][i]=1;    
    Mat ans=pow(a,p);
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=n;j++)
          cout<<ans.m[i][j]%mod<<" ";
        cout<<endl;
    }  
    return 0;
}

縮點（真題）

#include<iostream>
using namespace std;
const int maxn = 10007;
const int maxm = 100007;  
struct node
{
    int x,y;
}E[maxm],ne[maxm];
int n,m,tm = 0,top = 0,tot = 0;
int W[maxn],dfn[maxn],low[maxn],Head[maxn],Next[maxm],S[maxn],V[maxn],sd[maxn];//w 點權...sd 縮點後 舊點所對應的新點
int nh[maxn],nn[maxm],in[maxn],D[maxn];//newhead，newnext，index入度，D就是上面說的F啦
void tarjan(int x)
{
    dfn[x] = low[x] = ++tm;
    S[++top] = x; V[x] = 1;
    for (int y,i = Head[x]; i; i = Next[i])
    {
        y = E[i].y;
        if (!dfn[y])
        {
            tarjan(y);
            low[x] = min(low[x],low[y]);
        }
        else if (V[y])
        {
            low[x] = min(low[x],low[y]);// low[x] = min(low[x],dfn[y]);兩種都行，老師講的註釋裏的 求大佬解釋下到底哪一種好
        }
    }
    if (dfn[x] == low[x])
    {
        int y;
        while (y = S[top--])
        {
              V[y] = 0;
              sd[y] = x;//縮點
              if (y==x) break;
              W[x] +=W[y];
        }
    }
    return;
}
int main()
{
    cin>>n>>m;
    for (int i = 1; i<=n; i++)
    {
        scanf("%d",&W[i]);
    }
    for (int i = 1; i<=m; i++)
    {
        scanf("%d%d",&E[i].x,&E[i].y);
        Next[i] = Head[E[i].x];
        Head[E[i].x] = i;
    }    
    for (int i = 1; i<=n; i++)
    {
        if (!dfn[i])
        {
            tarjan(i);
        }
    }
    for (int x,y,i = 1; i<=m; i++)
    {
        x = sd[E[i].x]; y = sd[E[i].y];
        if (x!=y)//不是一個強連通份量 說明須要建邊
        {
            ne[++tot].x = x; ne[tot].y = y;
            nn[tot] = nh[x];
            nh[x] = tot;
            in[y]++;
        }
    }
    top =  0;
    for (int i = 1; i<=n; i++)
    {
        if (sd[i] == i && !in[i])//找那些入度爲零的強連通份量根節點
        {
            S[++top] = i;
            D[i] = W[i];
        }
    }
    while (top)
    {
        int x = S[top--];
        for (int y,i = nh[x]; i; i = nn[i])
        {
            y = ne[i].y;
            D[y] = max(D[y],D[x]+W[y]);
            if (--in[y] == 0)
            {
                S[++top] = y;
            }
        }
    }
    int ans = 0;
    for (int i = 1; i<=n; i++)
    {
        ans = max(ans,D[i]);
    }
    cout<<ans<<endl;
    return 0;
}

線性篩素數（真題）

#include<bits/stdc++.h>
using namespace std;
int n,m,x;
bool shai(int a)
{
    if(a==1) return 0;
    if(a==2 || a==3) return 1;
    if(a%6!=1 && a%6!=5) return 0;
    for(int i=5;i<=sqrt(a);i+=6)
    {
        if(a%i==0 || a%(i+2)==0)
            return 0;
    }
    return 1;
}
int main()
{
    cin>>n>>m;
    for(int i=1;i<=m;i++)
    {
        cin>>x;
        if(shai(x))
            cout<<"Yes"<<endl;
        else
            cout<<"No"<<endl;
        x=0;
    }
    return 0;
}

最近公共祖先(真題)

#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
int n,m,s,tot=0,cnt=0;
int head[1000100],nxt[1000100],to[1000100];
int d[500100],f[30][1000100];
int read()
{
    int x=0,flag=0;
    char ch=getchar();
    if(ch=='-') flag=1;
    while(ch<'0'||ch>'9')ch=getchar();
    while(ch>='0'&&ch<='9')x*=10,x+=ch-'0',ch=getchar();
    if(flag) return -x;
    return x;
}
void addedge(int x,int y)
{
    cnt++;
    nxt[cnt]=head[x];
    head[x]=cnt;
    to[cnt]=y;
}
void dfs(int u,int dep)
{
    d[u]=dep;
    for(int i=head[u];i!=-1;i=nxt[i])
    {
        int v=to[i];
        if(!d[v]) dfs(v,dep+1),f[0][v]=u;
    }
}
int LCA(int x,int y)
{
    int l=0;
    while((1<<l)<=n) l++;
    l--;
    if(d[x]<d[y]) swap(x, y);
    for(int i=20;i>=0;i--)
        if(d[y]<=d[x]-(1<<i)) x=f[i][x];
    if(x==y) return x;
    for(int i=20;i>=0;i--)
    {
        if(f[i][x]!=f[i][y])
        {
            x=f[i][x];
            y=f[i][y];
        }
    }
    return f[0][x];
}
int main()
{
    memset(head,-1,sizeof(head));
    n=read(),m=read(),s=read();
    for(int i=1;i<n;i++)
    {
        int x,y;
        x=read(),y=read();
        addedge(x,y);
        addedge(y,x);
    }
    f[0][s]=s;
    dfs(s,1);
    for(int i=1;(1<<i)<=n;i++)
        for(int j=1;j<=n;j++)
            f[i][j]=f[i-1][f[i-1][j]];
    for(int i=1;i<=m;i++)
    {
        int l,r;
        l=read(),r=read();
        printf("%d\n",LCA(l,r));
    }
    return 0;
}

最長公共子序列（真題）

#include<bits/stdc++.h> 
using namespace std;
int a[110000],t[110000],A[110000],B[110000],f[110000];
bool cmp(int a,int b)
{
    return a<b;
}
int solve(int l,int r,int x)
{
    int mid=(l+r)/2;;
    if (l==r) return l;
    if (a[mid]>x) 
        return solve(l,mid,x);
    if (a[mid]<=x) 
        return solve(mid+1,r,x);
}
int main()
{
    int n;
    scanf("%d",&n);
    for (int i=1; i<=n; i++) scanf("%d",&A[i]);
    for (int i=1; i<=n; i++) scanf("%d",&B[i]);
    for (int i=1; i<=n; i++) f[A[i]]=i;
    for (int i=1; i<=n; i++) t[i]=f[B[i]];
    memset(a,0,sizeof(a));
    for (int i=1; i<=n; i++)
    {
        if ((i==0)||(t[i]>a[a[0]])) 
            a[++a[0]]=t[i];
        else 
        if (t[i]<a[a[0]]) 
            a[solve(1,a[0],t[i])]=t[i];
    }
    printf("%d\n",a[0]);
    return 0;
}

最小生成樹（真題）

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
using namespace std;
const int M=200100;
struct node
{
    int x;
    int y;
    int w;
} a[M],t;
int n,e,dad[M],p=1,ans;

bool cmp(node x,node y)
{
    if(x.w<y.w)return 1;
    if(x.w==y.w)
        if(x.x>y.x)return 1;
    return 0;
}

void qsort(int l,int r)
{
    int i=l,j=r;
    node mid=a[(i+j)/2];
    while(i<=j)
    {
        while(cmp(a[i],mid))i++;
        while(cmp(mid,a[j]))j--;
        if(i<=j)
        {
            t=a[i];
            a[i]=a[j];
            a[j]=t;
            i++;
            j--;
        }
    }
    if(i<r)qsort(i,r);
    if(l<j)qsort(l,j);
}
int findx(int x)
{
    if(x==dad[x])return x;
    dad[x]=findx(dad[x]);
    return dad[x];
}
void solve()
{
    qsort(1,e);
    for(int i=1; i<=n; i++)dad[i]=i;
    for(int i=1; i<=e; i++)
    {
        if(findx(a[i].x)!=findx(a[i].y))
        {
            ans+=a[i].w;
            dad[findx(a[i].x)]=a[i].y;
            p++;
            if(p==n)return ;
        }
    }
}
int main()
{
    cin>>n>>e;
    for(int i=1; i<=e; i++)cin>>a[i].x>>a[i].y>>a[i].w;
    solve();
    cout<<ans;
    return 0;
}

快讀

long long read()
{
    long long x=0,f=1;
    char ch=getchar();
    while(ch<'0'||ch>'9')
    {
        if(ch=='-')f=-1;
        ch=getchar();
    }
    while(ch>='0'&&ch<='9')
    {
        x=x*10+ch-'0';
        ch=getchar();
    }
    return x*f;
}

嚴格次小生成樹（真題）

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>

using namespace std;
const int N = 1e5 + 10;

#define gc getchar()

int n, m, now = 1;
struct Node
{
    int u, v, w, is_in;
    bool operator <(const Node a)const
    {
        return w < a.w;
    }
} E[N * 3];
struct Node_2
{
    int u, v, w, nxt;
} G[(N * 3) << 1];
int fa[N][30], Max[N][30], Tmax[N][30], deep[N], head[N], father[N];
int Mi[30];

#define LL long long
LL Answer;
int Min = 1e9;

inline int read()
{
    int x = 0;
    char c = gc;
    while(c < '0' || c > '9') c = gc;
    while(c >= '0' || c >= '9') x = x * 10 + c - '0', c = gc;
    return x;
}

void Add(int u, int v, int w)
{
    G[now].v = v;
    G[now].w = w;
    G[now].nxt = head[u];
    head[u] = now ++;
}
int Getfa(int x)
{
    return father[x] == x ? x : father[x] = Getfa(father[x]);
}

void Mst()
{
    int js(0);
    for(int i = 1; js != n - 1; i ++)
    {
        int u = E[i].u, v = E[i].v, fu = Getfa(u), fv = Getfa(v);
        if(fu != fv)
        {
            father[fu] = fv;
            js ++;
            E[i].is_in = 1;
            Answer += (LL)E[i].w;
            Add(E[i].u, E[i].v, E[i].w);
            Add(E[i].v, E[i].u, E[i].w);
        }
    }
}

void Dfs(int x, int f_, int dep)
{
    deep[x] = dep;
    for(int i = 1; ; i ++)
    {
        if(deep[x] - Mi[i] < 0) break;
        fa[x][i] = fa[fa[x][i - 1]][i - 1];
        Max[x][i] = max(Max[x][i - 1], Max[fa[x][i - 1]][i - 1]);
        if(Max[x][i - 1] == Max[fa[x][i - 1]][i - 1])
            Tmax[x][i] = max(Tmax[x][i - 1], Tmax[fa[x][i - 1]][i - 1]);
        else
            Tmax[x][i] = max(min(Max[x][i - 1], Max[fa[x][i - 1]][i - 1]),
                             max(Tmax[x][i - 1], Tmax[fa[x][i - 1]][i - 1]));
    }
    for(int i = head[x]; ~ i; i = G[i].nxt)
    {
        int v = G[i].v;
        if(v != f_)
        {
            fa[v][0] = x;
            Max[v][0] = G[i].w;
            Tmax[v][0] = -1;
            Dfs(v, x, dep + 1);
        }
    }
}

int Lca(int x, int y)
{
    if(deep[x] < deep[y]) swap(x, y);
    int k = deep[x] - deep[y];
    for(int i = 0; i <=  17; i ++)
        if(k >> i & 1) x = fa[x][i];
    if(x == y) return x;
    for(int i = 17; i >= 0; i --)
        if(fa[x][i] != fa[y][i]) x = fa[x][i], y = fa[y][i];
    return fa[x][0];
}

void Work(int s, int t, int w_)
{
    int m1 = 0, m2 = 0, k = deep[s] - deep[t];
    for(int i = 0; i <= 17; i ++)
    {
        if(k >> i & 1)
        {
            m2 = max(m2, Tmax[s][i]);
            if(Max[s][i] > m1)
            {
                m2 = max(m2, m1);
                m1 = Max[s][i];
            }
        }
    }
    if(m1 == w_) Min = min(Min, w_ - m2);
    else Min = min(Min, w_ - m1);
}

int main()
{
    n = read();
    m = read();
    for(int i = 1; i <= n; i ++) head[i] = -1, father[i] = i;
    Mi[0] = 1;
    for(int i = 1; i <= 17; i ++) Mi[i] = Mi[i - 1] * 2;
    for(int i = 1; i <= m; i ++)
    {
        E[i].u = read(), E[i].v = read(), E[i].w = read();
    }
    sort(E + 1, E + m + 1);
    Mst();
    Dfs(1, 0, 1);
    for(int i = 1; i <= m; i ++)
    {
        if(!E[i].is_in)
        {
            int u = E[i].u, v = E[i].v;
            int L = Lca(u, v);
            Work(u, L, E[i].w);
            Work(v, L, E[i].w);
        }
    }
    cout << Answer + Min;
    return 0;
}

最大子段和（真題）

#include <stdlib.h>
#include<stdio.h>
int main()
{
    int count;
    int a[100];
    int b[100];
    int i;
    int max;
    scanf("%d",&count);
    for(i=0; i<count; i++)
    {
        scanf("%d",&a[i]);
    }
    b[0]=a[0];
    max=b[0];
    for(i=1; i<count; i++)
    {
        if(b[i-1]>0)
            b[i]=b[i-1]+a[i];
        else
            b[i]=a[i];
        if(b[i]>max)
            max=b[i];
    }
    printf("%d\n",max);
    return 0;
}

快速冪||取餘運算（真題）

#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
long long b,p,k,s,t;
int main()
{
    cin>>b>>p>>k;
    cout<<b<<"^"<<p<<" mod "<<k<<"=";
    s=b%k;
    t=1;
    for (int i=2;i<=p;i++)
    {
        s=s*b%k;
        if (s==b%k) break;
        t++;
    }
    p=p%t;s=1;
    if (p==0) p=t;
    for (int i=1;i<=p;i++)
    s=s*b%k;
    cout<<s;
    return 0;
}

快速冪

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
using namespace std;
int ksm(int a,int b,int p)
{
    int ans=1;
    while(b)
    {
        if(b&1)             //這一位是否是1 
            ans=ans*a%p;
        a=a*a%p;
        b=b>>1;             //送過來新的一位 
    }
    return ans;
}
int main()
{
    int a,b,p;
    cin>>a>>b>>p;
    cout<<ksm(a,b,p);
    return 0;
}

卡特蘭數

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
using namespace std;
int a[1000010];
int main()
{
    int n;
    cin>>n;
    a[0]=1;
    a[1]=1;
    for(int i=2;i<=n;i++)
    {
        a[i]=a[i-1]*(4*i-2)/(i+1);
    }
    cout<<a[n];
    return 0;
}

歐拉篩法

#include <cstring>
using namespace std;
int prime[1100000],primesize,phi[11000000];
bool is[11000000];
void getlist(int listsize)
{
    memset(is,1,sizeof(is));
    is[1]=false;
    for(int i=2; i<=listsize; i++)
    {
        if(is[i])
            prime[++primesize]=i;
        for(int j=1; j<=primesize&&i*prime[j]<=listsize; j++)
        {
            is[i*prime[j]]=false;
            if(i%prime[j]==0)break;
        }
    }
}

Dijkstra

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<queue>
using namespace std;
struct edge
{
    int u,v,x;
} ed[1000010];
struct node
{
    int x,y;
    bool operator <(const node & a)const
    {
        return y>a.y;
    }
};
priority_queue<node>que;
int head[1000010];
int dis[1000010];
int n,m,s,cnt;
void add(int x,int y,int z)
{
    cnt++;
    ed[cnt].u=head[x];
    ed[cnt].v=y;
    ed[cnt].x=z;
    head[x]=cnt;
}
void dij()
{
    for(int i=1; i<=n; i++)
        dis[i]=2147483647;
    dis[s]=0;
    que.push((node)
    {
        s,0
    });
    while(!que.empty())
    {
        node temp=que.top();
        que.pop();
        int x=temp.x,y=temp.y;
        if(y!=dis[x])
            continue;
        for(int i=head[x]; i; i=ed[i].u)
        {
            if(dis[ed[i].v]>dis[x]+ed[i].x)
            {
                dis[ed[i].v]=dis[x]+ed[i].x;
                que.push((node)
                {
                    ed[i].v,dis[ed[i].v]
                });
            }
        }
    }
}
int main()
{
    cin>>n>>m>>s;
    for(int i=1; i<=m; i++)
    {
        int x,y,z;
        cin>>x>>y>>z;
        add(x,y,z);
    }
    dij();
    for(int i=1; i<=n; i++)
        cout<<dis[i]<<" ";
    return 0;
}

單調隊列

#include <cstdio>
#include <algorithm>
using namespace std;

const int N = 100000;
int n, m, head = 1, tail = 0, ans = 2147483647;
int a[N + 1], f[N + 1], que[N + 1];

int main()
{
    freopen("beacon.in", "r", stdin);
    freopen("beacon.out", "w", stdout);

    scanf("%d%d", &n, &m);
    for (int i = 1;i <= n;i++)
        scanf("%d", &a[i]);
    for (int i = 1;i <= n;i++)
    {
        while (head <= tail && f[i - 1] <= f[que[tail]]) tail--; //當F[i-1]比隊尾值更優時把隊尾值彈出
        que[++tail] = i - 1; //把F[i-1]插入，這裏插入下標而不插入值，便於從隊頭彈出
        while (head <= tail && que[head] < i - m) head++; //不屬於區間維護內的數彈出
        f[i] = f[que[head]] + a[i]; //狀態轉移
    }
    for (int i = n;i > n - m;i--) //求出答案
        ans = min(ans, f[i]);
    printf("%d\n", ans);

    fclose(stdin);
    fclose(stdout);
    return 0;
}

EXGCD

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
using namespace std;
int exgcd(int a,int b,int &x,int &y)
{
    if(b==0)
    {
        x=1,y=0;
        return a;
    }
    int an=exgcd(b,a%b,x,y),temp=x;
    x=y,y=temp-a/b*y;
    return an;
}
int main()
{
    int a,b,x,y;
    cin>>a>>b;
    int ans=exgcd(a,b,x,y);
    while(x<0)
        x+=(b/ans);
    cout<<x;
    return 0;
}

KMP字符串匹配

#include<iostream>
#include<cstring>
#define MAXN 1000010
using namespace std;
int kmp[MAXN];
int la,lb,j; 
char a[MAXN],b[MAXN];
int main()
{
    cin>>a+1;
    cin>>b+1;
    la=strlen(a+1);
    lb=strlen(b+1);
    for (int i=2;i<=lb;i++)
       {     
       while(j&&b[i]!=b[j+1])
        j=kmp[j];    
       if(b[j+1]==b[i])j++;    
        kmp[i]=j;
       }
    j=0;
    for(int i=1;i<=la;i++)
       {
          while(j>0&&b[j+1]!=a[i])
           j=kmp[j];
          if (b[j+1]==a[i]) 
           j++;
          if (j==lb) {cout<<i-lb+1<<endl;j=kmp[j];}
       }

    for (int i=1;i<=lb;i++)
    cout<<kmp[i]<<" ";
    return 0;
}

字符串哈希

#include<bits/stdc++.h>
using namespace std;
typedef unsigned long long hh;
hh chang=131;
hh a[11000];
char s[11000];
int n,ans=1;
hh hashn(char s[])
{
    int len=strlen(s);
    hh ans=0;
    for(int i=0;i<len;i++)
        ans=ans*chang+(hh)s[i];
    return ans & 0x7fffffff;
}
int main()
{
    cin>>n;
    for(int i=1;i<=n;i++)
    {
        cin>>s;
        a[i]=hashn(s);
    }
    sort(a+1,a+n+1);
    for(int i=2;i<=n;i++)
        if(a[i]!=a[i-1])
        ans++;
    cout<<ans<<endl;
    return 0;
}

縮點（真題）

#include<bits/stdc++.h>
using namespace std;
const int N = 110000;
int dfn[N],low[N],head[N],color[N],sta[N],value[N];
int f[N],rd[N],dis[N],z[N],x[N],y[N];
int k=0,n,m,top=0,tim=0,ans=0,ct;
bool vis[N];
struct node
{
    int to,next,from;
}e[N];
void add(int u,int v)
{
    e[++k].to=v;
    e[k].from=u;
    e[k].next=head[u];
    head[u]=k;
}
void tarjan(int s)
{
    dfn[s]=low[s]=++tim;
    sta[++top]=s;
    vis[s]=true;
    for(int i=head[s],v=e[i].to;i;i=e[i].next,v=e[i].to)
    {
        if(!dfn[v])
            tarjan(v),low[s]=min(low[s],low[v]);
        else if(vis[v])
            low[s]=min(low[s],dfn[v]);
    }
    if(low[s]==dfn[s])
    {
        ++ct;
        vis[s]=false;
        while(sta[top+1]!=s)
        {
            color[sta[top]]=ct;
            z[ct]+=value[sta[top]];
            vis[sta[top]]=false;
            top--;
        }
    }
}
int dfs(int x)
{
    if(f[x])return f[x];
    f[x]=z[x];
    int maxx=0;
    for(int i=head[x];i;i=e[i].next)
    {
       int to=e[i].to;
       if(!f[to])dfs(to);
       if(maxx>f[to])
       maxx=maxx;
       else
       maxx=f[to];
    }
    f[x]+=maxx;
    return f[x];
}
int main()
{
    cin>>n>>m;
    for(int i=1;i<=n;i++) cin>>value[i];
    for(int i=1;i<=m;i++)
    {
        int u,v;
        cin>>u>>v;
        x[i]=u,y[i]=v;
        add(u,v);
    }
    for(int i=1;i<=n;i++)
        if(!dfn[i]) tarjan(i);
    memset(vis,false,sizeof(vis));
    memset(head,0,sizeof(head));
    memset(e,0,sizeof(e));
    k=0;
    for(int i=1;i<=m;i++)
    {
        if(color[x[i]]!=color[y[i]])
        {
            add(color[x[i]],color[y[i]]);
        }
    }
    for(int i=1;i<=ct;++i)
    if(!f[i])
    {
       dfs(i);
       if(ans>f[i])
       ans=ans;
       else 
       ans=f[i];
    }
    cout<<ans;
}

主席樹 （真題）

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 210000;
int n, m, root[MAXN], cut, a[MAXN], s[MAXN], t, x, y, z;
struct Edge
{
    int lc, rc, ans;
} tree[MAXN * 20];
void add(int &now, int last, int l, int r, int x)
{
    now = ++cut;
    tree[now].ans = tree[last].ans + 1;
    tree[now].lc = tree[last].lc, tree[now].rc = tree[last].rc;
    if(l == r) return ;
    int mid = (l + r) >> 1;
    if(x <= mid) add(tree[now].lc, tree[last].lc, l, mid, x);
    else add(tree[now].rc, tree[last].rc, mid + 1, r, x);
    return ;
}
int query(int L, int R, int l, int r, int x)
{
    if(l == r) return l;
    int p = tree[tree[R].lc].ans - tree[tree[L].lc].ans;
    int mid = (l + r) >> 1;
    if(p >= x) return query(tree[L].lc, tree[R].lc, l, mid, x);
    else return query(tree[L].rc, tree[R].rc, mid + 1, r, x - p);
}
int main()
{
    cin >> n >> m;
    for(int i = 1; i <= n; i++) cin >> s[i], a[i] = s[i];
    sort(s + 1, s + n + 1);
    for(int i = 1; i <= n; i++)
    {
        int p = lower_bound(s + 1, s + n + 1, a[i]) - s;
        add(root[i], root[i - 1], 1, n, p);
    }
    while(m--)
    {
        cin >> x >> y >> z;
        int p = query(root[x - 1], root[y], 1, n, z);
        printf("%d\n", s[p]);
    }
    return 0;
}

行來春色三分雨，睡去巫山一片雲