Time Limit: 5000/5000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 3301 Accepted Submission(s): 1552
Problem Description Network flow is a well-known difficult problem for ACMers. Given a graph, your task is to find out the maximum flow for the weighted directed graph.Input The first line of input contains an integer T, denoting the number of test cases. For each test case, the first line contains two integers N and M, denoting the number of vertexes and edges in the graph. (2 <= N <= 15, 0 <= M <= 1000) Next M lines, each line contains three integers X, Y and C, there is an edge from X to Y and the capacity of it is C. (1 <= X, Y <= N, 1 <= C <= 1000)
Output For each test cases, you should output the maximum flow from source 1 to sink N.
Sample Input 2 3 2 1 2 1 2 3 1 3 3 1 2 1 2 3 1 1 3 1
Sample Output Case 1: 1 Case 2: 2
Author HyperHexagon
Source HyperHexagon's Summer Gift (Original tasks)
Recommend zhengfeng 用最大流最短路径增广算法（EK算法）解决 #include<iostream> #include<queue> using namespace std; #define INF 0xfffffff #define maxn 50 #define min(x,y)(x<y?x:y) int cap[maxn][maxn]; int n,m,s,t,maxflow; int flow[maxn][maxn],parent[maxn],d[maxn]; void Edmonds_Karp(int s,int t,int nnum) { maxflow=0; memset(flow,0,sizeof(flow)); memset(parent,0,sizeof(parent)); while(true) { memset(d,0,sizeof(d)); queue<int>que; d[s]=INF; que.push(s); while(!que.empty()) { int u=que.front(); que.pop(); for(int v=0;v<nnum;v++) { if(!d[v]&&cap[u][v]>flow[u][v]) { parent[v]=u; que.push(v); d[v]=min(d[u],cap[u][v]-flow[u][v]); } } } if(!d[t]) break; for(int u=t;u!=s;u=parent[u]) { flow[parent[u]][u]+=d[t]; flow[u][parent[u]]-=d[t]; } maxflow+=d[t]; } } int main() { int t,cases=0,i; scanf("%d",&t); while(t--) { memset(cap,0,sizeof(cap)); scanf("%d %d",&n,&m); while(m--) { int x,y,c; scanf("%d %d %d",&x,&y,&c); cap[x-1][y-1]+=c; } Edmonds_Karp(0,n-1,n); printf("Case %d: %d\n",++cases,maxflow); } return 0; }
转载于:https://www.cnblogs.com/Su-Blog/archive/2012/08/14/2638841.html
