hdu 3549 Flow Problem （ek算法模板）

Flow Problem

Time Limit: 5000/5000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 7363    Accepted Submission(s): 3412

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

 

Author

HyperHexagon

#include"stdio.h"#include"string.h"#include"queue"using namespace std;#define N 20const int inf=10000000;int g[N][N];int pre[N],mark[N];int ek(int n){    int i,u,d,ans=0;    while(1)    {        queue<int>q;        q.push(1);        memset(mark,0,sizeof(mark));        memset(pre,0,sizeof(pre));        mark[1]=1;        while(!q.empty())        {            u=q.front();            q.pop();            for(i=1;i<=n;i++)            {                if(!mark[i]&&g[u][i])                {                    mark[i]=1;                    pre[i]=u;                    q.push(i);                }            }        }        if(pre[n]==0)            break;        d=inf;        for(i=n;i!=1;i=pre[i])        {            d=min(d,g[pre[i]][i]);        }        for(i=n;i!=1;i=pre[i])        {            g[pre[i]][i]-=d;            g[i][pre[i]]+=d;        }        ans+=d;    }    return ans;}int main(){    int n,m,T,u,v,w,cnt=1;    scanf("%d",&T);    while(T--)    {        memset(g,0,sizeof(g));        scanf("%d%d",&n,&m);        while(m--)        {            scanf("%d%d%d",&u,&v,&w);            g[u][v]+=w;        }        printf("Case %d: %d\n",cnt++,ek(n));    }    return 0;}