poj 3469 Dual Core CPU 最小割

题目链接

好裸的题.......

两个cpu分别作为源点和汇点， 每个cpu向元件连边， 权值为题目所给的两个值， 如果两个元件之间有关系， 就在这两个元件之间连边， 权值为消耗，这里的边应该是双向边。

 #include<bits/stdc++.h>
 using namespace std;
 #define mem(a) memset(a, 0, sizeof(a))
 const int maxn = 3e5+;
 int head[maxn*], s, t, num, q[maxn*], dis[maxn];
 struct node
 {
     int to, nextt, c;
 }e[maxn*];
 void init() {
     mem1(head);
     num = ;
 }
 void add(int u, int v, int c) {
     e[num].to = v; e[num].nextt = head[u]; e[num].c = c; head[u] = num++;
 }
 int bfs() {
     int u, v, st = , ed = ;
     mem(dis);
     dis[s] = ;
     q[ed++] = s;
     while(st<ed) {
         u = q[st++];
         for(int i = head[u]; ~i; i = e[i].nextt) {
             v = e[i].to;
             if(e[i].c&&!dis[v]) {
                 dis[v] = dis[u]+;
                 if(v == t)
                     return ;
                 q[ed++] = v;
             }
         }
     }
     return ;
 }
 int dfs(int u, int limit) {
     if(u == t)
         return limit;
     int cost = ;
     for(int i = head[u]; ~i; i = e[i].nextt) {
         int v = e[i].to;
         if(e[i].c&&dis[u] == dis[v]-) {
             int tmp = dfs(v, min(limit-cost, e[i].c));
             if(tmp>) {
                 e[i].c -= tmp;
                 e[i^].c += tmp;
                 cost += tmp;
                 if(cost == limit)
                     break;
             } else {
                 dis[v] = -;
             }
         }
     }
     return cost;
 }
 int dinic() {
     int ans = ;
     while(bfs()) {
         ans += dfs(s, inf);
     }
     return ans;
 }
 int main()
 {
     int n, m, x, y, z;
     while(~scanf("%d%d", &n, &m)) {
         s = , t = n+;
         init();
         for(int i = ; i<=n; i++) {
             scanf("%d%d", &x, &y);
             add(s, i, x);
             add(i, s, );
             add(i, t, y);
             add(t, i, );
         }
         while(m--) {
             scanf("%d%d%d", &x, &y, &z);
             add(x, y, z);
             add(y, x, z);
         }
         printf("%d\n", dinic());
     }
 }