分析

跟上一题差不多，把每个女生拆成两个点，如果男生女生之间可以配对就第一个点链接，不能配对就第二个点之间链接，最后两个点之间连一条流量是 k k k的边

代码

#include <iostream>
#include <cstring>

using namespace std;

const int N = 2010,M = N * N * 2;
const int INF = 0x3f3f3f3f;
int n,m,S,T,k;
int h[N],ne[M],e[M],f[M],idx;
int q[N], d[N], cur[N];
bool st[N][N];
int p[N];

template<typename T>inline void read(T &a){char c=getchar();T x=0,f=1;while(!isdigit(c)){if(c=='-')f=-1;c=getchar();}
while(isdigit(c)){x=(x<<1)+(x<<3)+c-'0';c=getchar();}a=f*x;}
int gcd(int a,int b){return (b>0)?gcd(b,a%b):a;}

void add(int x,int y,int z){
    e[idx] = y,ne[idx] = h[x],f[idx] = z,h[x] = idx++;
    e[idx] = x,ne[idx] = h[y],f[idx] = 0,h[y] = idx++;
}

bool bfs(){
    int hh = 0, tt = 0;
    memset(d,-1,sizeof d);
    q[0] = S, d[S] = 0, cur[S] = h[S];
    while(hh <= tt){
        int t = q[hh ++ ];
        for(int i = h[t];~i;i = ne[i]){
            int j = e[i];
            if(d[j] == -1 && f[i]){
                d[j] = d[t] + 1;
                cur[j] = h[j];
                if(j == T) return true;
                q[ ++ tt] = j;
            }
        }
    }
    return false;
}

int find(int u,int limit){
    if(u == T) return limit;
    int flow = 0;
    for(int i = cur[u];~i && flow < limit;i = ne[i]){
        cur[u] = i;
        int j = e[i];
        if(d[j] == d[u] + 1 && f[i]){
            int t = find(j,min(f[i],limit - flow));
            if(!t) d[j] = -1;
            f[i] -= t,f[i ^ 1] +=t,flow += t;
        }
    }
    return flow;
}

int dinic(){
    int res = 0,flow;
    while(bfs()) while(flow = find(S,INF)) res += flow;
    return res;
}
int find(int x){
    if(x != p[x]) p[x] = find(p[x]);
    return p[x];
}

bool check(int mid){
    memset(h,-1,sizeof h);
    idx = 0;
    for(int i = 1;i <= n;i++)
        for(int j = 1;j <= n;j++)
            if(st[i][j])
                add(i,j + n,1);
    for(int i = 1;i <= n;i++) add(S,i,mid);
    for(int i = 1;i <= n;i++) add(i + n,T,mid);
    if(dinic() != mid * n) return false;
    return true;
}

int main(){
    int ssss;
    read(ssss);
    while(ssss--){
        memset(st,0,sizeof st);
        read(n),read(m),read(k);
        for(int i = 1;i <= n;i++) p[i] = i;
        S = 0,T = n * 2 + 1;
        while(m--){
            int x,y;
            read(x),read(y);
            st[x][y] = 1;
        }
        while(k--){
            int x,y;
            read(x),read(y);
            x = find(x),y = find(y);
            p[x] = y;
        }
        for(int i = 1;i <= n;i++)
            for(int j = 1;j <= n;j++){
                if(i == j || find(i) != find(j)) continue;
                for(int t = 1;t <= n;t++)
                    if(st[i][t]) st[j][t] = 1;
            }
        int l = 0,r = INF;
        while(l < r){
            int mid = (l + r + 1) >> 1;
            if(check(mid)) l = mid;
            else r = mid - 1;
        }
        printf("%d\n",l);
    }
    return 0;
}