Flow Problem

Time Limit: 5000/5000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 23036    Accepted Submission(s): 10756

 

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

 問題連結：HDU3549 Flow Problem

問題描述：給你一個N個頂點M條邊的有向圖,要你求1號點到N號點的最大流。

解題思路：最大流模板題，使用EdmondsKarp演算法（鄰接矩陣形式）

Source

HyperHexagon's Summer Gift (Original tasks)

AC的C++程式：

#include<iostream>
#include<cstring>
#include<algorithm>
#include<queue>

using namespace std;

const int N=20;
const int INF=0x3f3f3f3f;

int g[N][N];
int pre[N];
bool vis[N];

int n,m;

int EdmondsKarp()
{
	queue<int>q;
	memset(pre,-1,sizeof(pre));
	memset(vis,false,sizeof(vis));
	vis[1]=true;
	pre[1]=0;
	q.push(1);
	int v;
	bool flag=false;
	while(!q.empty())
	{
		v=q.front();
		q.pop();
		for(int i=1;i<=n;i++)
		  if(g[v][i]>0&&!vis[i])
		  {
			  pre[i]=v;		  	
			  vis[i]=true;
		  	  if(i==n)
		  	  {
		  	      flag=true;
		  	      break;
			  }
			  else
			    q.push(i);
		  }
	}
	if(!flag)//找不到就返回 
	  return 0;
	int ans=INF;
	for(v=n;v!=1;v=pre[v])
	  ans=min(ans,g[pre[v]][v]);
	for(v=n;v!=1;v=pre[v])
	{
		g[pre[v]][v]-=ans;
		g[v][pre[v]]+=ans;
	} 
	return ans;
}

int main()
{
	int T;
	scanf("%d",&T);
	for(int t=1;t<=T;t++)
	{
		scanf("%d%d",&n,&m);
		int a,b,c;
		memset(g,0,sizeof(g));
		for(int i=1;i<=m;i++)
		{
			scanf("%d%d%d",&a,&b,&c);
			g[a][b]+=c;
		}
		int ans=0,temp;
		while(temp=EdmondsKarp())
		  ans+=temp;
		printf("Case %d: %d\n",t,ans);
	}
	return 0;
}