【AtCoder】AGC013

AGC013

A - Sorted Arrays

直接分就行

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define MAXN 100005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    while(c < '0' || c > '9') {
    if(c == '-') f = -1;
    c = getchar();
    }
    while(c >= '0' && c <= '9') {
    res = res * 10 +c - '0';
    c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
    out(x / 10);
    }
    putchar('0' + x % 10);
}
int N;
int a[MAXN];
void Solve() {
    read(N);
    for(int i = 1 ; i <= N ; ++i) read(a[i]);
    int f = 0;
    int ans = N;
    for(int i = 2 ; i <= N ; ++i) {
    if(!f) {
        if(a[i] > a[i - 1]) f = 1;
        else if(a[i] < a[i - 1]) f = -1;
        --ans;
    }
    else {
        if(a[i] > a[i - 1] && f == 1) --ans;
        else if(a[i] < a[i - 1] && f == -1) --ans;
        else if(a[i] == a[i - 1]) --ans;
        else f = 0;
    }
    }
    out(ans);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}

B - Hamiltonish Path

直接一條路擴展到不能擴展位置

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define MAXN 100005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    while(c < '0' || c > '9') {
    if(c == '-') f = -1;
    c = getchar();
    }
    while(c >= '0' && c <= '9') {
    res = res * 10 +c - '0';
    c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
    out(x / 10);
    }
    putchar('0' + x % 10);
}
struct node {
    int to,next;
}E[MAXN * 2];
int N,M,head[MAXN],sumE;
int que[MAXN * 2],ql,qr;
bool vis[MAXN];
void add(int u,int v) {
    E[++sumE].to = v;
    E[sumE].next = head[u];
    head[u] = sumE;
}
void Solve() {
    read(N);read(M);
    int a,b;
    for(int i = 1 ; i <= M ; ++i) {
    read(a);read(b);
    add(a,b);add(b,a);
    }
    ql = N,qr = N - 1;
    que[++qr] = 1;vis[1] = 1;
    while(1) {
    int u = que[qr];
    bool flag = 1;
    for(int i = head[u] ; i ; i = E[i].next) {
        int v = E[i].to;
        if(!vis[v]) {
        que[++qr] = v;
        vis[v] = 1;
        flag = 0;break;
        }
    }
    if(flag) break;
    }
    while(1) {
    int u = que[ql];
    bool flag = 1;
    for(int i = head[u] ; i ; i = E[i].next) {
        int v = E[i].to;
        if(!vis[v]) {
        que[--ql] = v;
        vis[v] = 1;
        flag = 0;break;
        }
    }
    if(flag) break;
    }
    out(qr - ql + 1);enter;
    for(int i = ql ; i <= qr ; ++i) {
    out(que[i]);space;
    }
    enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}

C - Ants on a Circle

顯然能夠求出全部螞蟻最後的位置，如今要肯定第一我的的位置

逆時針轉，撞一下編號減一，順時針轉，撞一下編號加一，模擬一下便可

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define MAXN 100005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    while(c < '0' || c > '9') {
    if(c == '-') f = -1;
    c = getchar();
    }
    while(c >= '0' && c <= '9') {
    res = res * 10 +c - '0';
    c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
    out(x / 10);
    }
    putchar('0' + x % 10);
}
int N;
int64 T,x[MAXN],to[MAXN],ans[MAXN],L;
int id[MAXN],w[MAXN];
void Solve() {
    read(N);read(L);read(T);
    for(int i = 0 ; i < N ; ++i) {
    read(x[i]);read(w[i]);
    if(w[i] == 1) to[i] = (x[i] + T) % L;
    else to[i] = ((x[i] - T) % L + L) % L;
    id[i] = i;
    }
    sort(id,id + N,[](int a,int b){return to[a] < to[b] || (to[a] == to[b] && w[a] < w[b]);});
    int pos = 0;
    for(int i = 1 ; i < N ; ++i) {
    if(w[i] ^ w[0]) {
        pos += (T / L) * 2 % N;
        int64 rem = T % L;
        int64 dis = x[i] - x[0];
        if(w[0] == 2) dis = L - dis;
        if(dis + L < 2 * rem) pos += 2;
        else if(dis < 2 * rem) pos += 1;
        pos %= N;
    }
    }
    if(w[0] == 1) pos %= N;
    else pos = (N - pos % N) % N;
    for(int i = 0 ; i < N ; ++i) {
    if(id[i] == 0) {
        int p = i;
        int cnt = 0;
        while(cnt < N) {
        ans[pos] = to[id[p]];
        p = (p + 1) % N;
        pos = (pos + 1) % N;
        ++cnt;
        }
    }
    }
    for(int i = 0 ; i < N ; ++i) {
    out(ans[i]);enter;
    }
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}

D - Piling Up

假如如今盒子裏有\(x\)個紅球，\(N - x\)個藍球

\(dp[i][x]\)表示前\(2i\)個放完以後盒子裏有\(x\)個紅球

而後有四種

\(RR\)，要求\(x > 0\)，而後x - 1

\(RB\)，要求\(x > 0\)，而後x不變

\(BB\)，要求\(x < N\)，而後x + 1

\(BR\)，要求\(x < N\)，而後x不變

爲了避免重複統計，能夠直接設置一個臨近底部的特殊標記，若是這個操做序列畫出的x變化圖像不能下移才記錄

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define MAXN 100005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    while(c < '0' || c > '9') {
    if(c == '-') f = -1;
    c = getchar();
    }
    while(c >= '0' && c <= '9') {
    res = res * 10 +c - '0';
    c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
    out(x / 10);
    }
    putchar('0' + x % 10);
}
int MOD = 1000000007;
int dp[3005][3005][2];
int N,M;
int inc(int a,int b) {
    return a + b >= MOD ? a + b - MOD : a + b;
}
int mul(int a,int b) {
    return 1LL * a * b % MOD;
}
void update(int &x,int y) {
    x = inc(x,y);
}
void Solve() {
    read(N);read(M);
    for(int i = 0 ; i <= N ; ++i) dp[0][i][0] = 1;
    for(int i = 0 ; i < M ; ++i) {
    for(int j = 0 ; j <= N ; ++j) {
        for(int k = 0 ; k <= 1 ; ++k) {
        if(!dp[i][j][k]) continue;
        if(j >= 1) {
            int t = 0;
            if(j == 1) t = 1;
            update(dp[i + 1][j - 1][k | t],dp[i][j][k]);
            update(dp[i + 1][j][k | t],dp[i][j][k]);
        }
        if(j <= N - 1) {
            int t = 0;
            if(j == 0) t = 1;
            update(dp[i + 1][j + 1][k | t],dp[i][j][k]);
            update(dp[i + 1][j][k | t],dp[i][j][k]);
        }
        }
    }
    }
    int ans = 0;
    for(int j = 0 ; j <= N ; ++j) update(ans,dp[M][j][1]);
    out(ans);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}

E - Placing Squares

平方和轉有序對

直接統計\(dp[x][i = 0,1,2]\)爲從上一個端點開始選了幾個點的方案

能夠矩陣乘法，特殊端點處停下便可

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define MAXN 100005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    while(c < '0' || c > '9') {
    if(c == '-') f = -1;
    c = getchar();
    }
    while(c >= '0' && c <= '9') {
    res = res * 10 +c - '0';
    c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
    out(x / 10);
    }
    putchar('0' + x % 10);
}

int MOD = 1000000007;
int inc(int a,int b) {
    return a + b >= MOD ? a + b - MOD : a + b;
}
int mul(int a,int b) {
    return 1LL * a * b % MOD;
}
void update(int &x,int y) {
    x = inc(x,y);
}
struct Matrix {
    int f[3][3];
    Matrix() {memset(f,0,sizeof(f));}
    friend Matrix operator * (const Matrix &a,const Matrix &b) {
    Matrix c;
    for(int i = 0 ; i < 3 ; ++i) {
        for(int j = 0 ; j < 3 ; ++j) {
        for(int k = 0 ; k < 3 ; ++k) {
            update(c.f[i][j],mul(a.f[i][k],b.f[k][j]));
        }
        }
    }
    return c;
    }
    void unit() {
    for(int i = 0 ; i < 3 ; ++i) f[i][i] = 1;
    }
}A,B,ans;
int N,M,x[MAXN];
Matrix fpow(Matrix a,int c) {
    Matrix res,t = a;
    res.unit();
    while(c) {
    if(c & 1) res = res * t;
    t = t * t;
    c >>= 1;
    }
    return res;
}
void Solve() {
    read(N);read(M);
    ans.unit();
    A.f[0][1] = 2;A.f[0][2] = 1;A.f[0][0] = 1;
    A.f[1][2] = 1;A.f[1][1] = 1;
    A.f[2][2] = 2;A.f[2][0] = 1;A.f[2][1] = 2;
    B = A;B.f[2][0] = 0;B.f[2][1] = 0;B.f[2][2] = 1;
    int p = 0;
    for(int i = 1 ; i <= M ; ++i) {
    read(x[i]);
    ans = ans * fpow(A,x[i] - p);
    ans = ans * B;
    p = x[i] + 1;
    }
    ans = ans * fpow(A,N - p);
    out(ans.f[0][2]);enter;
    
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}

F - Two Faced Cards

離散化以後把\([C[i],tot]\)減1

而後把正面的\([A[i],tot]\)加1

而後序列中不能有小於0的數，由於咱們最後還能加上一張卡，在這以前咱們須要知足全部的負值大於等於-1

咱們從後往前掃不合法的負值，而後配上一個覆蓋這個點左端點最小的區間，這個必定要從右往左作，由於以後給了一張牌是能覆蓋一個後綴區間，以後的前綴區間須要另外的操做，咱們但願咱們把這個負值合法化的同時，前面的負值處理的最多

以後就是前綴區間須要補充的操做，對於每一個遇到負值元素選當前覆蓋它的區間右端點最大的區間便可

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define MAXN 100005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    while(c < '0' || c > '9') {
    if(c == '-') f = -1;
    c = getchar();
    }
    while(c >= '0' && c <= '9') {
    res = res * 10 +c - '0';
    c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
    out(x / 10);
    }
    putchar('0' + x % 10);
}
int N,Q,pos;
int A[MAXN][2],C[MAXN],ans;
int val[MAXN * 4],tot;
int cnt[MAXN * 4],num[MAXN * 4],f[MAXN * 4],inx[MAXN * 4];
bool all_fail;
struct cmp1 {
    bool operator () (pii a,pii b) {
    if(a.fi != b.fi) return a.fi < b.fi;
    return a.se > b.se;
    }
};
struct cmp2 {
    bool operator () (pii a,pii b) {
    if(a.se != b.se) return a.se > b.se;
    return a.fi < b.fi;
    }
};
multiset<pii,cmp1> S1;
multiset<pii,cmp2> S2;
vector<int> v[MAXN * 4];
vector<pii > rec;
void Init() {
    read(N);
    for(int i = 1 ; i <= N ; ++i) {
    read(A[i][0]);read(A[i][1]);
    val[++tot] = A[i][0];
    val[++tot] = A[i][1];
    }
    for(int i = 1 ; i <= N + 1 ; ++i) {
    read(C[i]);
    val[++tot] = C[i];
    }
    sort(val + 1,val + tot + 1);
    tot = unique(val + 1,val + tot + 1) - val - 1;
    for(int i = 1 ; i <= N ; ++i) {
    int t = lower_bound(val + 1,val + tot + 1,A[i][0]) - val;
    cnt[t]++;
    }
    for(int i = 1 ; i <= N + 1 ; ++i) {
    int t = lower_bound(val + 1,val + tot + 1,C[i]) - val;
    cnt[t]--;
    }
    for(int i = 1 ; i <= tot ; ++i) {
    num[i] = cnt[i];
    cnt[i] += cnt[i - 1];
    }
    ans = N;
    for(int i = 1 ; i <= N ; ++i) {
    if(A[i][0] > A[i][1]) {
        int s = lower_bound(val + 1,val + tot + 1,A[i][0]) - val;
        int t = lower_bound(val + 1,val + tot + 1,A[i][1]) - val;
        v[s - 1].pb(t);
    }
    }
    int pre = 0;
    memset(inx,0,sizeof(inx));
    for(int i = tot ; i >= 1 ; --i) {
    for(auto s : v[i]) S1.insert(mp(s,i));
    pre += inx[i];
    cnt[i] += pre;
    while(cnt[i] < -1 && S1.size() > 0) {
        pii seg = *S1.begin();S1.erase(S1.begin());
        if(seg.fi > i) {rec.pb(seg);continue;}
        num[seg.fi]++;num[seg.se + 1]--;
        inx[seg.fi - 1]--;
        --ans;
        cnt[i]++;++pre;
    }
    if(cnt[i] < -1) all_fail = 1;
    }
    for(int i = 1 ; i <= tot ; ++i) {
    num[i] += num[i - 1];
    v[i].clear();
    }
    for(auto t : S1) {
    rec.pb(t);
    }
    for(auto t : rec) {
    v[t.fi].pb(t.se);
    }
    memset(inx,0,sizeof(inx));
    pre = 0;
    for(int i = 1 ; i <= tot ; ++i) {
    for(auto t : v[i]) S2.insert(mp(i,t));
    pre += inx[i];
    num[i] += pre;
    if(num[i] == -1) {
        if(S2.size() <= 0) {
        for(int j = i ; j <= tot ; ++j) f[j] = -1;
        break;
        }
        pii seg = *S2.begin();
        if(seg.se < i) {
        for(int j = i ; j <= tot ; ++j) f[j] = -1;
        break;
        }
        inx[seg.se + 1]--;
        ++pre;
        f[i] = f[i - 1] + 1;
    }
    else f[i] = f[i - 1];
    }
    f[tot + 1] = -1;
}

void Solve() {
    int Q;
    read(Q);
    int d,e;
    for(int i = 1 ; i <= Q ; ++i) {
    read(d);read(e);
    if(all_fail) {puts("-1");continue;}
    int res = -1;
    int t = lower_bound(val + 1,val + tot + 1,d) - val - 1;
    if(val[t - 1] == d - 1) {
        if(f[t - 1] != -1) res = max(ans - f[t - 1] + 1,res);
    }
    if(f[t] != -1) res = max(ans - f[t] + 1,res);
    t = lower_bound(val + 1,val + tot + 1,e) - val - 1;
    if(val[t - 1] == e - 1) {
        if(f[t - 1] != -1) res = max(ans - f[t - 1],res);
    }
    if(f[t] != -1) res = max(ans - f[t],res);
    out(res);enter;
    }
    
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Init();
    Solve();
}