POJ 1459：Power Network（最大流）

http://poj.org/problem?id=1459

题意：有np个发电站，nc个消费者，m条边，边有容量限制，发电站有产能上限，消费者有需求上限问最大流量。

思路：S和发电站相连，边权是产能上限，消费者和T相连，边权是需求上限，边的话就按题意加就好了。难点更觉得在于输入。。加个空格。。边数组要*2，因为有反向边。

 #include <cstdio>
 #include <algorithm>
 #include <iostream>
 #include <cstring>
 #include <string>
 #include <cmath>
 #include <queue>
 #include <vector>
 #include <map>
 #include <set>
 using namespace std;
 #define INF 0x3f3f3f3f
 #define N 110
 typedef long long LL;
 struct Edge {
     int v, cap, nxt;
     Edge () {}
     Edge (int v, int cap, int nxt) : v(v), cap(cap), nxt(nxt) {}
 }edge[N*N*];
 int tot, cur[N], dis[N], pre[N], head[N], gap[N], S, T;

 void Add(int u, int v, int cap) {
     edge[tot] = Edge(v, cap, head[u]); head[u] = tot++;
     edge[tot] = Edge(u, , head[v]); head[v] = tot++;
 }

 void BFS() {
     queue<int> que;
     while(!que.empty()) que.pop();
     memset(dis, -, sizeof(dis));
     memset(gap, , sizeof(gap));
     dis[T] = ; gap[]++; que.push(T);
     while(!que.empty()) {
         int u = que.front(); que.pop();
         for(int i = head[u]; ~i; i = edge[i].nxt) {
             int v = edge[i].v;
             if(dis[v] == -) continue;
             dis[v] = dis[u] + ;
             gap[dis[v]]++;
             que.push(v);
         }
     }
 }

 int ISAP(int n) {
     memcpy(cur, head, sizeof(cur));
     BFS();
     int u = pre[S] = S, ans = , i;
     while(dis[S] < n) {
         if(u == T) {
             int flow = INF, index;
             for(int i = S; i != T; i = edge[cur[i]].v) {
                 Edge& e = edge[cur[i]];
                 if(e.cap < flow) {
                     flow = e.cap; index = i;
                 }
             }
             for(int i = S; i != T; i = edge[cur[i]].v) {
                 edge[cur[i]].cap -= flow;
                 edge[cur[i]^].cap += flow;
             }
             ans += flow; u = index;
         }
         for(i = cur[u]; ~i; i = edge[i].nxt)
             if(dis[edge[i].v] == dis[u] -  && edge[i].cap > )
                 break;
         if(~i) {
             cur[u] = i;
             pre[edge[i].v] = u;
             u = edge[i].v;
         } else {
             int md = n;
             if(--gap[dis[u]] == ) break;
             for(int i = head[u]; ~i; i = edge[i].nxt) {
                 if(md > dis[edge[i].v] && edge[i].cap > ) {
                     md = dis[edge[i].v]; cur[u] = i;
                 }
             }
             ++gap[dis[u] = md + ];
             u = pre[u];
         }
     }
     return ans;
 }

 int main() {
     int n, nc, np, m;
     while(~scanf("%d%d%d%d", &n, &np, &nc, &m)) {
         tot = ; memset(head, -, sizeof(head));
         S = n, T = n + ;
         int a, b, c;
         for(int i = ; i <= m; i++) {
             scanf(" (%d,%d)%d", &a, &b, &c);
             Add(a, b, c);
         }
         for(int i = ; i <= np; i++) {
             scanf(" (%d)%d", &a, &b);
             Add(S, a, b);
         }
         for(int i = ; i <= nc; i++) {
             scanf(" (%d)%d", &a, &b);
             Add(a, T, b);
         }
         printf("%d\n", ISAP(T+));
     }
     return ;
 }