HDU 2549 Flow Problem

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

 

Source

HyperHexagon's Summer Gift (Original tasks)

 

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最大流~

还是模板题~

#include<cstdio>
#include<cstring>
#include<iostream>
#include<vector>
#include<queue>
using namespace std;

int t,n,m,x,y,v,a[1001],ne[1001],ans,cnt;

struct edge{
	int x,y,cap,flow;
	edge(int u,int v,int k,int z):x(u),y(v),cap(k),flow(z) {}
};

vector<edge> ed;
vector<int> p[2001];

int main()
{
	scanf("%d",&t);
	while(t--)
	{
		ans=0;ed.clear();
		for(int i=1;i<=n;i++) p[i].clear();
		scanf("%d%d",&n,&m);
		while(m--)
		{
			scanf("%d%d%d",&x,&y,&v);
			ed.push_back(edge(x,y,v,0));
			ed.push_back(edge(y,x,0,0));
			int tot=ed.size();
			p[x].push_back(tot-2);
			p[y].push_back(tot-1);
		}
		while(1)
		{
			memset(a,0,sizeof(a));
			queue<int> q;q.push(1);a[1]=999999999;
			while(!q.empty())
			{
				int k=q.front();q.pop();int tot=p[k].size();
				for(int i=0;i<tot;i++)
				{
					edge z=ed[p[k][i]];
					if(!a[z.y] && z.cap>z.flow)
					{
						a[z.y]=min(a[k],z.cap-z.flow);
						ne[z.y]=p[k][i];q.push(z.y);
					}
				}
				if(a[n]) break;
			}
			if(!a[n]) break;
			for(int i=n;i!=1;i=ed[ne[i]].x)
			{
				ed[ne[i]].flow+=a[n];
				ed[ne[i]^1].flow-=a[n];
			}
			ans+=a[n];
		}
		printf("Case %d: %d\n",++cnt,ans);
	}
	return 0;
}