hdu 3549 Flow Problem【最大流问题】【FF】

来源：互联网发布：网络运维与管理pdf 编辑：程序博客网时间：2024/06/03 22:40

Flow Problem

Time Limit: 5000/5000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 11848 Accepted Submission(s): 5627

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

23 21 2 12 3 13 31 2 12 3 11 3 1

Sample Output

Case 1: 1Case 2: 2

最大流入门问题，这里有个人对其理解的部分的详解：http://blog.csdn.net/mengxiang000000/article/details/50601356

直接上代码：

#include <stdio.h>#include <iostream>#include <string.h>#include <queue>using namespace std;const int N=1005;int pre[N];       //保存增广路径上的点的前驱顶点bool vis[N];int map[N][N];int s,t;          //s为源点，t为汇点int n,m;bool bfs(){    int i,cur;    queue<int >q;    memset(pre,0,sizeof(pre));    memset(vis,0,sizeof(vis));    vis[s]=true;    q.push(s);    while(!q.empty())    {        cur=q.front();        q.pop();        if(cur==t)return true;        for(int i=1;i<=n;i++)        {            if(vis[i]==0&&map[cur][i])            {                q.push(i);                pre[i]=cur;                vis[i]=true;            }        }    }    return false;}int Max_Flow(){    int ans=0;    while(1)    {        if(!bfs())return ans;        int Min=0x3f3f3f3f;        for(int i=t;i!=s;i=pre[i])        {            Min=min(Min,map[pre[i]][i]);        }        for(int i=t;i!=s;i=pre[i])        {            map[pre[i]][i]-=Min;            map[i][pre[i]]+=Min;        }        ans+=Min;    }}int main(){    int T,k=1;    int u,v,c;    scanf("%d",&T);    while(T--)    {        scanf("%d%d",&n,&m);        s=1;        t=n;        memset(map,0,sizeof(map));        for(int i=0;i<m;i++)        {            scanf("%d%d%d",&u,&v,&c);            map[u][v]+=c;        }        printf("Case %d: %d\n",k++,Max_Flow());    }}

0 0