【HDU3488】【匹配】【KM最小费用圈】

Tour

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)

Total Submission(s): 2324 Accepted Submission(s): 1166



Problem Description

In the kingdom of Henryy, there are N (2 <= N <= 200) cities, with M (M <= 30000) one-way roads connecting them. You are lucky enough to have a chance to have a tour in the kingdom. The route should be designed as: The route should contain one or more loops.
(A loop is a route like: A->B->……->P->A.)

Every city should be just in one route.

A loop should have at least two cities. In one route, each city should be visited just once. (The only exception is that the first and the last city should be the same and this city is visited twice.)

The total distance the N roads you have chosen should be minimized.



Input

An integer T in the first line indicates the number of the test cases.

In each test case, the first line contains two integers N and M, indicating the number of the cities and the one-way roads. Then M lines followed, each line has three integers U, V and W (0 < W <= 10000), indicating that there is a road from U to V, with the
distance of W.

It is guaranteed that at least one valid arrangement of the tour is existed.

A blank line is followed after each test case.



Output

For each test case, output a line with exactly one integer, which is the minimum total distance.



Sample Input

1
6 9
1 2 5
2 3 5
3 1 10
3 4 12
4 1 8
4 6 11
5 4 7
5 6 9
6 5 4

Sample Output

42

Source

2010 ACM-ICPC Multi-University
Training Contest（6）——Host by BIT



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#include <iostream>
#include <cstring>
#include <cmath>
#include <queue>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <string>
#include <cstdlib>
#include <cstdio>
#include <algorithm>
using namespace std;

const int N = 220;
const int INF = 0x3f3f3f3f;
int nx,ny;//两边的点数  
int g

;//二分图描述  
int linker
,lx
,ly
;//y中各点匹配状态，x,y中的点标号  
int slack
;  
bool visx
,visy
;  
  
bool DFS(int x)  
{  
    visx[x] = true;  
    for(int y = 0; y < ny; y++)  
    {  
        if(visy[y])continue;  
        int tmp = lx[x] + ly[y] - g[x][y];  
        if(tmp == 0)  
        {  
            visy[y] = true;  
            if(linker[y] == -1 || DFS(linker[y]))  
            {  
                linker[y] = x;  
                return true;  
            }  
        }  
        else if(slack[y] > tmp)  
            slack[y] = tmp;  
    }  
    return false;  
}  
int KM()  
{  
    memset(linker,-1,sizeof(linker));  
    memset(ly,0,sizeof(ly));  
    for(int i = 0;i < nx;i++)  
    {  
        lx[i] = -INF;  
        for(int j = 0;j < ny;j++)  
            if(g[i][j] > lx[i])  
                lx[i] = g[i][j];  
    }  
    for(int x = 0;x < nx;x++)  
    {  
        for(int i = 0;i < ny;i++)  
            slack[i] = INF;  
        while(true)  
        {  
            memset(visx,false,sizeof(visx));  
            memset(visy,false,sizeof(visy));  
            if(DFS(x))break;  
            int d = INF;  
            for(int i = 0;i < ny;i++)  
                if(!visy[i] && d > slack[i])  
                    d = slack[i];  
            for(int i = 0;i < nx;i++)  
                if(visx[i])  
                    lx[i] -= d;  
            for(int i = 0;i < ny;i++)  
            {  
                if(visy[i])ly[i] += d;  
                else slack[i] -= d;  
            }  
        }  
    }  
    int res = 0;  
    for(int i = 0;i < ny;i++)  
        if(linker[i] != -1)  
            res += g[linker[i]][i];  
    return res;  
}  
int T;
int main()
{
    scanf("%d",&T);
	while(T --)
	{
		int n,m;
		memset(g,~INF,sizeof(g));
		scanf("%d%d",&n,&m);
		for(int i=0;i<m;i++)
		{
			int u,v,w;
			scanf("%d%d%d",&u,&v,&w);
			u --;
			v --;
			g[u][v] = max(g[u][v],-w);
		}
		nx = n;
		ny = n;
		printf("%d\n",-KM());
	}
    return 0;
}