【AtCoder】ARC074

ARC 074

C - Chocolate Bar

直接枚舉第一刀橫切豎切，而後另外一塊要求若是橫切分紅\(H / 2\)豎切分紅\(W/2\)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 200005
//#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);
}
int64 H,W,ans = 1e18;
int64 maxdec(int64 a,int64 b,int64 c) {
    return max(max(abs(a - b),abs(a - c)),abs(b - c));
}
void Solve() {
    read(H);read(W);
    for(int i = 1 ; i < H ; ++i) {
    int64 h = (H - i) / 2;
    ans = min(ans,maxdec(i * W,h * W,(H - i - h) * W));
    int64 t = W / 2;
    ans = min(ans,maxdec(i * W,t * (H - i),(W - t) * (H - i)));
    }
    for(int i = 1 ; i < W ; ++i) {
    int64 h = (W - i) / 2;
    ans = min(ans,maxdec(i * H,h * H,(W - i - h) * H));
    int64 t = H / 2;
    ans = min(ans,maxdec(i * H,t * (W - i),(H - t) * (W - i)));
    }
    out(ans);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}

D - 3N Numbers

直接算前i個數最大的N個是多少，後i個數最小的是多少，能夠用set

#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 300005
//#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 cmp {
    bool operator () (const int &a,const int &b) const {
    return a > b;
    }
};
int N,a[MAXN];
int64 f[MAXN],b[MAXN];
multiset<int> s2;
multiset<int,cmp> s1;
void Solve() {
    read(N);
    for(int i = 1 ; i <= 3 * N ; ++i) read(a[i]);
    int64 sum = 0;
    for(int i = 1 ; i <= 3 * N ; ++i) {
    s1.insert(a[i]);sum += a[i];
    if(s1.size() > N) {
        auto t = *(--s1.end());
        s1.erase(--s1.end());
        sum -= t;
    }
    f[i] = sum;
    }
    sum = 0;
    for(int i = 3 * N ; i >= 1 ; --i) {
    s2.insert(a[i]);sum += a[i];
    if(s2.size() > N) {
        auto t = *(--s2.end());
        s2.erase(--s2.end());
        sum -= t;
    }
    b[i] = sum;
    }
    int64 ans = f[N] - b[N + 1];
    for(int i = N + 1 ; i <= 2 * N ; ++i) {
    ans = max(ans,f[i] - b[i + 1]);
    }
    out(ans);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}

E - RGB Sequence

就是記錄\(dp[r][g][b]\)爲上一次出現r的位置，出現g的位置和出現b的位置，把不合法的狀態標成0

#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 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);
}
const int MOD = 1000000007;
int N,M;
int dp[305][305][305],sum[3][305][305];
vector<pii > v[305];
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;
}
bool check(int r,int g,int b,int len) {
    for(int i = 0 ; i < v[len].size() ; ++i) {
    pii t = v[len][i];
    int cnt = 0;
    if(r >= t.fi) ++cnt;
    if(g >= t.fi) ++cnt;
    if(b >= t.fi) ++cnt;
    if(cnt != t.se) return false;
    }
    return true;
}
void increase(int r,int g,int b) {
    sum[0][g][b] = inc(sum[0][g][b],dp[r][g][b]);
    sum[1][r][b] = inc(sum[1][r][b],dp[r][g][b]);
    sum[2][r][g] = inc(sum[2][r][g],dp[r][g][b]);
}
void Solve() {
    read(N);read(M);
    int l,r,x;
    for(int i = 1 ; i <= M ; ++i) {
    read(l);read(r);read(x);
    v[r].pb(mp(l,x));
    }
    dp[0][0][0] = 1;
    sum[0][0][0] = 1;
    sum[1][0][0] = 1;
    sum[2][0][0] = 1;
    for(int i = 1 ; i <= N ; ++i) {
    for(int j = 0 ; j < i ; ++j) {
        for(int k = 0 ; k < i ; ++k) {
        dp[i][j][k] = sum[0][j][k];
        dp[j][i][k] = sum[1][j][k];
        dp[j][k][i] = sum[2][j][k];
        if(!check(i,j,k,i)) dp[i][j][k] = 0;
        if(!check(j,i,k,i)) dp[j][i][k] = 0;
        if(!check(j,k,i,i)) dp[j][k][i] = 0;
        }
    }
    memset(sum,0,sizeof(sum));
    for(int j = 0 ; j < i ; ++j) {
        for(int k = 0 ; k < i ; ++k) {
        increase(i,j,k);
        increase(j,i,k);
        increase(j,k,i);
        }
    }
    }
    int ans = 0;
    for(int i = 0 ; i < N ; ++i) {
    for(int j = 0 ; j < N ; ++j) {
        ans = inc(ans,dp[N][i][j]);
        ans = inc(ans,dp[i][N][j]);
        ans = inc(ans,dp[i][j][N]);
    }
    }
    out(ans);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}

F - Lotus Leaves

行列拆點，對於而後每一個葉子新建一個點往對應的行列連邊

對於起點和重點的邊建成無窮大

而後跑網絡流求最小割便可

#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 200005
//#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,cap;
}E[MAXN];
int sumE = 1,head[MAXN];
int H,W,Ncnt,S,T;
char s[105][105];
void add(int u,int v,int c) {
    E[++sumE].to = v;
    E[sumE].next = head[u];
    E[sumE].cap = c;
    head[u] = sumE;
}
void addtwo(int u,int v,int c) {
    add(u,v,c);add(v,u,0);
}
int gap[MAXN],dis[MAXN];
int isap(int u,int aug) {
    if(u == T) return aug;
    int flow = 0;
    for(int i = head[u] ; i ;i = E[i].next) {
    int v = E[i].to;
    if(E[i].cap) {
        if(dis[v] + 1 == dis[u]) {
        
        int t = isap(v,min(aug - flow,E[i].cap));
        flow += t;
        E[i].cap -= t;
        E[i ^ 1].cap += t;
        if(aug == flow) return flow;//這塊寫錯了，若是能滿流的話應該返回而不是讓dis++
        if(dis[S] >= Ncnt) return flow;
        }
    } 
    }
   
    --gap[dis[u]];
    if(!gap[dis[u]]) {dis[S] = Ncnt;return flow;}
    dis[u]++;
    ++gap[dis[u]];
    return flow;
}
void Solve() {
    read(H);read(W);
    for(int i = 1 ; i <= H ; ++i) {
    scanf("%s",s[i] + 1);
    }
    Ncnt = H + W;
    pii p[2];
    for(int i = 1 ; i <= H ; ++i) {
    for(int j = 1 ; j <= W ; ++j) {
        if(s[i][j] != '.') {
        ++Ncnt;
        int t = 0;
        if(s[i][j] == 'o') t = 1;
        else t = 0x7fffffff;
        addtwo(Ncnt,i,t);addtwo(i,Ncnt,t);
        addtwo(j + H,Ncnt,t);addtwo(Ncnt,j + H,t);
        if(s[i][j] == 'S') {p[0] = mp(i,j);S = Ncnt;}
        if(s[i][j] == 'T') {p[1] = mp(i,j);T = Ncnt;}
        }
    }
    }
    if(p[0].fi == p[1].fi || p[0].se == p[1].se) {puts("-1");return;}
    int ans = 0;
    while(dis[S] < Ncnt) ans += isap(S,0x7fffffff);
    out(ans);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}