Tour HDU - 3488（最大权值匹配）

Tour

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.

InputAn 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.OutputFor 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

The route should contain one or more loops.

一个或多个环。。二分匹配足以。。  把权值取反 然后套最大权值匹配即可

注意有重边。。但我们是要最小值  取反后在输入的时候只保留max的值即可

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#include <cmath>
#define mem(a, b) memset(a, b, sizeof(a))
using namespace std;
const int maxn = 220, INF = 0x7fffffff;
int usedx[maxn], usedy[maxn], cy[maxn], cx[maxn], w[maxn][maxn], bx[maxn], by[maxn];
int nx, ny, n, minn, min_val, m;
int dfs(int u)
{
    usedx[u] =1 ;
    for(int i=1; i<=ny; i++)
    {
        if(usedy[i] == -1)
        {
            int t = bx[u] + by[i] - w[u][i];
            if(t == 0)
            {
                usedy[i] = 1;
                if(cy[i] == -1 || dfs(cy[i]))
                {
                    cy[i] = u;
                    cx[u] = i;
                    return 1;
                }
            }
            else if(t > 0)
                minn = min(minn, t);
        }
    }
    return 0;
}

void km()
{
    mem(cy, -1);
    mem(cx, -1);
    for(int i=0; i<=nx; i++) bx[i] = -INF;
    mem(by, 0);
    for(int i=1; i<=nx; i++)
        for(int j=1; j<=ny; j++)
                bx[i] = max(bx[i], w[i][j]);
    for(int i=1; i<=nx; i++)
    {
        while(1)
        {
            minn = INF;
            mem(usedx, -1);
            mem(usedy, -1);
            if(dfs(i)) break;
            for(int j=1; j<=nx; j++)
                if(usedx[j] != -1) bx[j] -= minn;
            for(int j=1; j<=ny; j++)
                if(usedy[j] != -1) by[j] += minn;
        }
    }
    min_val = 0;
    for(int i=1; i<=nx; i++)
        if(cx[i] != -1)
            min_val += w[i][cx[i]];
    printf("%d\n",-min_val);
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        for(int i=1; i<=n; i++)
            for(int j=1; j<=n; j++) w[i][j] = -INF;
        for(int i=1; i<=m; i++)
        {
            int u, v, c;
            scanf("%d%d%d",&u,&v,&c);
                w[u][v] = max(w[u][v], -c);
        }
        nx = ny = n;
        km();
    }
    return 0;
}

原文地址：https://www.cnblogs.com/WTSRUVF/p/9311819.html