题目链接：http://poj.org/problem?id=1459
题意：
总共有N个点，M条边，有np个点是加油站即源点，有nc个点是消耗点，即汇点。
给出M条边的容量，每个源点最大的流出量和每个汇点的最大流入量。
求最多可以消耗多少。
思路：
多源多汇网络流就是增加一个超级源点和超级汇点。
把超级源点向每个源点连一条边，容量为该源点流出的最大量。
把每个汇点向超级汇点连一条边，容量为该汇点规定的最大流入量。
然后求最大流。
1 //#include <bits/stdc++.h> 2 #include <iostream> 3 #include <cstring> 4 #include <cstdio> 5 #include <vector> 6 #include <queue> 7 using namespace std; 8 int n, m, s, t; 9 #define maxn 110 10 #define inf 0x3f3f3f3f 11 struct Edge 12 { 13 int from, to, cap, flow; 14 Edge(int f, int t, int c, int fl) 15 { 16 from = f; to = t; cap = c; flow = fl; 17 } 18 }; 19 vector <Edge> edges; 20 vector <int> G[maxn]; 21 int d[maxn], vis[maxn], cur[maxn]; 22 void AddEdge(int from, int to, int cap) 23 { 24 edges.push_back(Edge(from, to, cap, 0)); 25 edges.push_back(Edge(to, from, 0, 0)); 26 m = edges.size(); 27 G[from].push_back(m-2); 28 G[to].push_back(m-1); 29 } 30 bool bfs() 31 { 32 memset(vis, 0, sizeof(vis)); 33 d[s] = 0; 34 vis[s] = 1; 35 queue <int> q; 36 q.push(s); 37 while(!q.empty()) 38 { 39 int x = q.front(); q.pop(); 40 for(int i = 0; i < G[x].size(); i++) 41 { 42 Edge &e = edges[G[x][i]]; 43 if(!vis[e.to] && e.cap > e.flow) 44 { 45 d[e.to] = d[x] + 1; 46 vis[e.to] = 1; 47 q.push(e.to); 48 } 49 } 50 } 51 return vis[t]; 52 } 53 int dfs(int x, int a) 54 { 55 if(x == t || a == 0) return a; 56 int flow = 0, f; 57 for(int &i = cur[x]; i < G[x].size(); i++) 58 { 59 Edge &e = edges[G[x][i]]; 60 if(d[e.to] == d[x] + 1 && (f = dfs(e.to, min(a, e.cap - e.flow))) > 0) 61 { 62 e.flow += f; 63 edges[G[x][i]^1].flow -= f; 64 flow += f; 65 a -= f; 66 if(a == 0) break; 67 } 68 } 69 return flow; 70 } 71 int maxflow() 72 { 73 int flow = 0; 74 while(bfs()) 75 { 76 memset(cur, 0, sizeof(cur)); 77 flow += dfs(s, inf); 78 } 79 return flow; 80 } 81 int N, np, nc, M; 82 int main() 83 { 84 //freopen("in.txt", "r", stdin); 85 //freopen("out.txt", "w", stdout); 86 while(~scanf("%d%d%d%d", &N, &np, &nc, &M)) 87 { 88 edges.clear(); 89 for(int i = 0; i <= N+1; i++) G[i].clear(); 90 int a, b, c; char op; 91 for(int i = 1; i <= M; i++) 92 { 93 cin>>op; scanf("%d", &a); 94 cin>>op; scanf("%d", &b); 95 cin>>op; scanf("%d", &c); 96 AddEdge(a, b, c); 97 } 98 s = N; t = N+1; 99 for(int i = 1; i <= np; i++) 100 { 101 cin>>op; scanf("%d", &a); 102 cin>>op; scanf("%d", &c); 103 AddEdge(s, a, c); 104 } 105 for(int i = 1; i <= nc; i++) 106 { 107 cin>>op; scanf("%d", &a); 108 cin>>op; scanf("%d", &c); 109 AddEdge(a, t, c); 110 } 111 n = N+2; 112 int flow = maxflow(); 113 printf("%d ", flow); 114 } 115 return 0; 116 }