hdu3549 Flow Problem

                                           Flow Problem

Time Limit: 5000/5000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 7867    Accepted Submission(s): 3661

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
 
最大流开通的第二题，模板只要稍微改动一下就AC了，一遍AC，1——n的最大流问题，这里是t组测试数据，不断的输入即可。
注释写的很明白，不再在此赘述。
代码如下：
#include <algorithm>#include <iostream>#include <cstring>#include <cstdlib>#include <cstdio>#include <queue>#define inf 88888888using namespace std;int cap[200][200];int flow[200][200];int a[200];int main(){    int n,m;    int i,j;    int si,ei,ci;    int t;    int k=1;    while(scanf("%d",&t)!=EOF)        //while(scanf("%d %d",&n,&m)!=EOF)    {        while(t--)        {            scanf("%d %d",&n,&m);            memset(cap,0,sizeof(cap));            for(i=0; i<m; i++)            {                scanf("%d %d %d",&si,&ei,&ci);                cap[si][ei]+=ci;            }            queue<int>q;            memset(flow,0,sizeof(flow));            int f=0;            int p[210];            while(1)            {                memset(a,0,sizeof(a));                a[1]=inf;                q.push(1);                while(!q.empty())//BFS找增光链                {                    int u=q.front();                    q.pop();                    for(int v=1; v<=n; v++)                    {                        if(!a[v]&&cap[u][v]>flow[u][v])//找到新节点v                        {                            p[v]=u;//记录v的父亲                            q.push(v);//并加入队列                            a[v] = a[u]<cap[u][v]-flow[u][v]?a[u]:cap[u][v]-flow[u][v];                            //a[v]=a[u]<?cap[u][v]-flow[u][v];//s-v路径上的最小残量                        }                    }                }                if(a[n]==0) break;//找不到，则当前流已经是最大流                for(int u=n; u!=1; u=p[u])//从汇点往回走                {                    flow[p[u]][u]+=a[n];//更新正向流量                    flow[u][p[u]]-=a[n];//更新反向流量                }                f+=a[n];//更新从s流出的总流量            }            printf("Case %d: %d\n",k++,f);//最后打印总的流量        }    }    return 0;}