题意：给一个无向图，每条边上都有容量的限制，要求求出给定起点和终点的最大流。
思路：每条无向边就得拆成2条，每条还得有反向边，所以共4条。源点汇点已经给出，所以不用建了。直接在图上跑最大流就可以了。
1 #include <bits/stdc++.h> 2 #define LL long long 3 #define pii pair<int,int> 4 #define INF 0x7f7f7f7f 5 using namespace std; 6 const int N=200; 7 const int mod=1e9+7; 8 int s, t; 9 10 int path[N], flow[N]; 11 vector<int> vect[N]; 12 13 struct node 14 { 15 int from, to, cap, flow; 16 node(){}; 17 node(int from,int to,int cap,int flow):from(from),to(to),cap(cap),flow(flow){}; 18 }edge[100000]; 19 int edge_cnt; 20 21 void add_node(int from,int to,int cap,int flow) 22 { 23 edge[edge_cnt]=node(from,to,cap,flow); 24 vect[from].push_back(edge_cnt++); 25 } 26 27 int BFS(int s,int e) 28 { 29 deque<int> que(1,s); 30 flow[s]=INF; 31 while(!que.empty()) 32 { 33 int x=que.front(); 34 que.pop_front(); 35 for(int i=0; i<vect[x].size(); i++) 36 { 37 node e=edge[vect[x][i]]; 38 if(!flow[e.to] && e.cap>e.flow) 39 { 40 flow[e.to]=min(flow[e.from],e.cap-e.flow); 41 path[e.to]=vect[x][i]; 42 que.push_back(e.to); 43 } 44 } 45 if(flow[e]) return flow[e]; 46 } 47 return flow[e]; 48 } 49 50 int max_flow(int s,int e) 51 { 52 int ans_flow=0; 53 while(true) 54 { 55 memset(path,0,sizeof(path)); 56 memset(flow,0,sizeof(flow)); 57 58 int tmp=BFS(s,e); 59 if(!tmp) return ans_flow; 60 ans_flow+=tmp; 61 62 int ed=e; 63 while(ed!=s) 64 { 65 int t=path[ed]; 66 edge[t].flow+=tmp; 67 edge[t^1].flow-=tmp; 68 ed=edge[t].from; 69 } 70 } 71 } 72 int main() 73 { 74 freopen("input.txt", "r", stdin); 75 int n, a, b, v, c, j=0; 76 while(scanf("%d",&n),n) 77 { 78 edge_cnt=0; 79 memset(edge,0,sizeof(edge)); 80 for(int i=0; i<=n+1; i++) vect[i].clear(); 81 82 scanf("%d%d%d", &s, &t, &c); 83 for(int i=0; i<c; i++) 84 { 85 scanf("%d%d%d",&a,&b,&v); 86 add_node(a, b, v, 0); 87 add_node(b, a, 0, 0); 88 add_node(b, a, v, 0); 89 add_node(a, b, 0, 0); 90 } 91 printf("Network %d ",++j); 92 printf("The bandwidth is %d. ", max_flow(s ,t) ); 93 } 94 return 0; 95 }