Flow Problem

TimeLimit:5000MS MemoryLimit:32768KB

64-bit integer IO format:%I64d

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.

SampleInput

SampleOutput

Case 1: 1
Case 2: 2
题解：模板题，求最大流量

  1 #include<bits/stdc++.h>
  2 using namespace std;
  3 const int maxn=1000;
  4 const int inf=0x3f3f3f3f;
  5 struct node
  6 {
  7     int to,flow,next;
  8 } edge[maxn*4];
  9 int first[maxn],sign,vis[maxn],pre[maxn];
 10 void init()
 11 {
 12     memset(first,-1,sizeof(first));
 13     sign=0;
 14 }
 15 void add_edge(int u,int v,int flow)
 16 {
 17     edge[sign].to=v;
 18     edge[sign].flow=flow;
 19     edge[sign].next=first[u];
 20     first[u]=sign++;
 21     edge[sign].to=u;
 22     edge[sign].flow=0;///建一条反向边，流量为0；
 23     edge[sign].next=first[v];
 24     first[v]=sign++;
 25 }
 26 bool bfs(int s,int t)
 27 {
 28     memset(vis,0,sizeof(vis));
 29     memset(pre,-1,sizeof(pre));
 30        vis[s]=1;
 31        queue<int >que;
 32        que.push(s);
 33        while(!que.empty())
 34        {
 35            int now=que.front();
 36            que.pop();
 37            if(now==t)
 38            {
 39                return 1;
 40            }
 41            for(int i=first[now];~i;i=edge[i].next)
 42            {
 43                int to=edge[i].to,flow=edge[i].flow;
 44                if(!vis[to]&&flow>0)
 45                {
 46                    vis[to]=1;
 47                    que.push(to);
 48                    pre[to]=i;///记录路径  记录的是由上一条边到to这个点
 49                }
 50            }
 51        }
 52        return 0;
 53 
 54 }
 55 int edomon_krap(int s,int t)///起点 终点
 56 {
 57     int max_flow=0;
 58     while(bfs(s,t))
 59     {
 60         int min_flow=inf;
 61         int x=t;
 62         while(x!=s)
 63         {
 64            // printf("%d
",x);
 65             int index=pre[x];
 66             //printf("intdex  %d
",index);
 67             min_flow=min(min_flow,edge[index].flow);
 68             x=edge[index^1].to;
 69              //printf("x==%d
",x);
 70         }
 71          x=t;
 72         while(x!=s)
 73         {
 74             int index=pre[x];
 75             edge[index].flow-=min_flow;
 76             edge[index^1].flow+=min_flow;///反向边加上min_flow
 77             x=edge[index^1].to;
 78         }
 79         max_flow+=min_flow;
 80 
 81     }
 82    return max_flow;
 83 }
 84 int main()
 85 {
 86     int t,n,m;
 87     scanf("%d",&t);
 88     for(int i=1;i<=t;i++)
 89     {
 90         init();
 91         scanf("%d%d",&n,&m);
 92         for(int j=1;j<=m;j++)
 93         {
 94             int u,v,w;
 95             scanf("%d%d%d",&u,&v,&w);
 96             add_edge(u,v,w);
 97         }
 98         printf("Case %d: %d
",i,edomon_krap(1,n));
 99     }
100 }