HDUOJ 3488 Tour

HDUOJ 3488 Tour

題目鏈接

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

比較巧妙的二分圖最大權匹配問題~
有人可能會問，這不是求最小距離嗎？我們只要將所有邊權值取負即可，初始化將所有邊權值賦 -$inf$，此時問題就轉化爲二分圖最大權匹配問題了，套一個 $KM$ 模板即可，注意答案最後取負輸出，AC代碼如下：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=210;
const int inf=1e8;
int n,m,w[N][N],lx[N],ly[N],visx[N],visy[N],slack[N],link[N];
bool found(int x){
    visx[x]=1;
    for(int y=1;y<=n;y++){
        if(visy[y]) continue;
        int t=lx[x]+ly[y]-w[x][y];
        if(t==0){
            visy[y]=1;
            if(link[y]==0 || found(link[y])){
                link[y]=x;
                return 1;
            }
        }
        else if(slack[y]>t) slack[y]=t;
    }
    return 0;
}

void KM(){
    fill(lx,lx+N,0);
    fill(ly,ly+N,0);
    fill(link,link+N,0);
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++)
            if(lx[i]<w[i][j]) lx[i]=w[i][j];
    }
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++) slack[j]=inf;
        while(1){
            fill(visx,visx+N,0);
            fill(visy,visy+N,0);
            if(found(i)) break;
            int d=inf;
            for(int k=1;k<=n;k++)
                if(!visy[k] && d>slack[k]) d=slack[k];
            for(int k=1;k<=n;k++){
                if(visx[k]) lx[k]-=d;
                if(visy[k]) ly[k]+=d;
            }
        }
    }
    int ans=0;
    for(int i=1;i<=n;i++)
        if(w[link[i]][i]!=inf) ans+=w[link[i]][i];
    printf("%d\n",-ans);
}

int main(){
    int t;
    cin>>t;
    while(t--){
        scanf("%d%d",&n,&m);
        fill(w[0],w[0]+N*N,-inf);
        int x,y,z;
        for(int i=1;i<=m;i++){
            scanf("%d%d%d",&x,&y,&z);
            w[x][y]=max(w[x][y],-z);
        }
        KM();
    }
    return 0;
}