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：

  1 #include<bits/stdc++.h>
  2 using namespace std;
  3 #define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);
  4 #define LL long long
  5 #define ULL unsigned LL
  6 #define fi first
  7 #define se second
  8 #define pb emplace_back
  9 #define lson l,m,rt<<1
 10 #define rson m+1,r,rt<<1|1
 11 #define lch(x) tr[x].son[0]
 12 #define rch(x) tr[x].son[1]
 13 #define max3(a,b,c) max(a,max(b,c))
 14 #define min3(a,b,c) min(a,min(b,c))
 15 typedef pair<int,int> pll;
 16 const int inf = 0x3f3f3f3f;
 17 const LL INF = 0x3f3f3f3f3f3f3f3f;
 18 const LL mod =  (int)1e9+7;
 19 const int N = 300;
 20 const int M = N * N;
 21 int head[N], deep[N], cur[N];
 22 int w[M], to[M], nx[M];
 23 int tot;
 24 void add(int u, int v, int val){
 25     w[tot]  = val; to[tot] = v;
 26     nx[tot] = head[u]; head[u] = tot++;
 27
 28     w[tot] = 0; to[tot] = u;
 29     nx[tot] = head[v]; head[v] = tot++;
 30 }
 31 int bfs(int s, int t){
 32     queue<int> q;
 33     memset(deep, 0, sizeof(deep));
 34     q.push(s);
 35     deep[s] = 1;
 36     while(!q.empty()){
 37         int u = q.front();
 38         q.pop();
 39         for(int i = head[u]; ~i; i = nx[i]){
 40             if(w[i] > 0 && deep[to[i]] == 0){
 41                 deep[to[i]] = deep[u] + 1;
 42                 q.push(to[i]);
 43             }
 44         }
 45     }
 46     return deep[t] > 0;
 47 }
 48 int Dfs(int u, int t, int flow){
 49     if(u == t) return flow;
 50     for(int &i = cur[u]; ~i; i = nx[i]){
 51         if(deep[u]+1 == deep[to[i]] && w[i] > 0){
 52             int di = Dfs(to[i], t, min(w[i], flow));
 53             if(di > 0){
 54                 w[i] -= di, w[i^1] += di;
 55                 return di;
 56             }
 57         }
 58     }
 59     return 0;
 60 }
 61
 62 int Dinic(int s, int t){
 63     int ans = 0, tmp;
 64     while(bfs(s, t)){
 65         for(int i = 0; i <= t; i++) cur[i] = head[i];
 66         while(tmp = Dfs(s, t, inf)) ans += tmp;
 67     }
 68     return ans;
 69 }
 70
 71 void init(){
 72     memset(head, -1, sizeof(head));
 73     tot = 0;
 74 }
 75 int pre[N];
 76 int link[N][N];
 77 int edge[M][2];
 78 int Find(int x){
 79     if(x == pre[x]) return x;
 80     return pre[x] = Find(pre[x]);
 81 }
 82 int T, n, m, f, s, t, u, v;
 83 void GG(int val){
 84     for(int i = head[s]; ~i; i = nx[i]){
 85         if(i&1);
 86         else w[i] = val, w[i+1] = 0;
 87     }
 88     for(int i = n+1; i <= n+n; i++){
 89         for(int j = head[i]; ~j; j = nx[j]){
 90             if(j&1);
 91             else w[j] = val, w[j+1] = 0;
 92         }
 93     }
 94     for(int i = 1; i <= n; i++){
 95         for(int j = head[i]; ~j; j = nx[j]){
 96             if(j&1);
 97             else w[j] = 1, w[j+1] = 0;
 98         }
 99     }
100 }
101 int main(){
102     scanf("%d", &T);
103     while(T--){
104         init();
105         scanf("%d%d%d", &n, &m, &f);
106         for(int i = 1; i <= n; i++){
107             pre[i] = i;
108             for(int j = 1; j <= n; j++)
109                 link[i][j] = 0;
110         }
111         for(int i = 1; i <= m; i++)
112             scanf("%d%d", &edge[i][0], &edge[i][1]);
113         for(int i = 1; i <= f; i++){
114             scanf("%d%d", &u, &v);
115             u = Find(u);
116             v = Find(v);
117             if(v == u) continue;
118             pre[u] = v;
119         }
120         for(int i = 1; i <= m; i++){
121             int z = Find(edge[i][0]);
122             link[z][edge[i][1]] = 1;
123         }
124         s = 0, t = 2 * n + 1;
125         for(int i = 1; i <= n; i++){
126             add(s, i, 1);
127             int z = Find(i);
128             for(int j = 1; j <= n; j++){
129                 if(link[z][j])
130                     add(i, j+n, 1);
131             }
132             add(i+n,t,1);
133         }
134         int l = 1, r = n;
135         while(l <= r){
136             int mid = l+r >> 1;
137             GG(mid);
138             if(Dinic(s,t) == mid*n) l = mid + 1;
139             else r = mid - 1;
140         }
141         printf("%d\n", l-1);
142     }
143     return 0;
144 }

原文地址：https://www.cnblogs.com/MingSD/p/9738729.html