HDU 3081 Marriage Match II 二分 + 网络流

Marriage Match II

题意：有n个男生，n个女生，现在有 f 条男生女生是朋友的关系， 现在有 m 条女生女生是朋友的关系， 朋友的朋友是朋友，现在进行 k 轮游戏，每轮游戏都要男生和女生配对，每轮配对过的人在接下来中都不能配对，求这个k最大是多少。

题解：二分 + 网络流check 。

代码：

 #include<bits/stdc++.h>
 using namespace std;
 #define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);
 #define LL long long
 #define ULL unsigned LL
 #define fi first
 #define se second
 #define pb emplace_back
 #define lson l,m,rt<<1
 #define rson m+1,r,rt<<1|1
 #define lch(x) tr[x].son[0]
 #define rch(x) tr[x].son[1]
 #define max3(a,b,c) max(a,max(b,c))
 #define min3(a,b,c) min(a,min(b,c))
 typedef pair<int,int> pll;
 const int inf = 0x3f3f3f3f;
 const LL INF = 0x3f3f3f3f3f3f3f3f;
 const LL mod =  (int)1e9+;
 const int N = ;
 const int M = N * N;
 int head[N], deep[N], cur[N];
 int w[M], to[M], nx[M];
 int tot;
 void add(int u, int v, int val){
     w[tot]  = val; to[tot] = v;
     nx[tot] = head[u]; head[u] = tot++;

     w[tot] = ; to[tot] = u;
     nx[tot] = head[v]; head[v] = tot++;
 }
 int bfs(int s, int t){
     queue<int> q;
     memset(deep, , sizeof(deep));
     q.push(s);
     deep[s] = ;
     while(!q.empty()){
         int u = q.front();
         q.pop();
         for(int i = head[u]; ~i; i = nx[i]){
             if(w[i] >  && deep[to[i]] == ){
                 deep[to[i]] = deep[u] + ;
                 q.push(to[i]);
             }
         }
     }
     return deep[t] > ;
 }
 int Dfs(int u, int t, int flow){
     if(u == t) return flow;
     for(int &i = cur[u]; ~i; i = nx[i]){
         if(deep[u]+ == deep[to[i]] && w[i] > ){
             int di = Dfs(to[i], t, min(w[i], flow));
             if(di > ){
                 w[i] -= di, w[i^] += di;
                 return di;
             }
         }
     }
     return ;
 }

 int Dinic(int s, int t){
     int ans = , tmp;
     while(bfs(s, t)){
         for(int i = ; i <= t; i++) cur[i] = head[i];
         while(tmp = Dfs(s, t, inf)) ans += tmp;
     }
     return ans;
 }

 void init(){
     memset(head, -, sizeof(head));
     tot = ;
 }
 int pre[N];
 int link[N][N];
 int edge[M][];
 int Find(int x){
     if(x == pre[x]) return x;
     return pre[x] = Find(pre[x]);
 }
 int T, n, m, f, s, t, u, v;
 void GG(int val){
     for(int i = head[s]; ~i; i = nx[i]){
         if(i&);
         else w[i] = val, w[i+] = ;
     }
     for(int i = n+; i <= n+n; i++){
         for(int j = head[i]; ~j; j = nx[j]){
             if(j&);
             else w[j] = val, w[j+] = ;
         }
     }
     for(int i = ; i <= n; i++){
         for(int j = head[i]; ~j; j = nx[j]){
             if(j&);
             else w[j] = , w[j+] = ;
         }
     }
 }
 int main(){
     scanf("%d", &T);
     while(T--){
         init();
         scanf("%d%d%d", &n, &m, &f);
         for(int i = ; i <= n; i++){
             pre[i] = i;
             for(int j = ; j <= n; j++)
                 link[i][j] = ;
         }
         for(int i = ; i <= m; i++)
             scanf("%d%d", &edge[i][], &edge[i][]);
         for(int i = ; i <= f; i++){
             scanf("%d%d", &u, &v);
             u = Find(u);
             v = Find(v);
             if(v == u) continue;
             pre[u] = v;
         }
         for(int i = ; i <= m; i++){
             int z = Find(edge[i][]);
             link[z][edge[i][]] = ;
         }
         s = , t =  * n + ;
         for(int i = ; i <= n; i++){
             add(s, i, );
             int z = Find(i);
             for(int j = ; j <= n; j++){
                 if(link[z][j])
                     add(i, j+n, );
             }
             add(i+n,t,);
         }
         int l = , r = n;
         while(l <= r){
             int mid = l+r >> ;
             GG(mid);
             if(Dinic(s,t) == mid*n) l = mid + ;
             else r = mid - ;
         }
         printf("%d\n", l-);
     }
     return ;
 }