HDU 3488 Tour (KM)

Tour

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 1266    Accepted Submission(s): 666

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

 

 

Recommend

zhouzeyong

 

 

 

题目大体意思是求n个环的并，每个点只能在一个环里，找到可以遍历所有顶点的边的最小值，显然每个点只能关联其他两个顶点，并且入度和初度都为1，所以应该是完全匹配，完全匹配图就是n个环的并，所以求的是完美匹配中的最小权问题，用km解决即可~

 

KM算法详解见：http://www.cnblogs.com/jackge/archive/2013/05/03/3057028.html

 

#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;

const int VM=250;
const int INF=0x3f3f3f3f;

int n,m,nx,ny;
int linker[VM],lx[VM],ly[VM],slack[VM];
int visx[VM],visy[VM],w[VM][VM];

int DFS(int x){
    visx[x]=1;
    for(int y=1;y<=ny;y++){
        if(visy[y])
            continue;
        int tmp=lx[x]+ly[y]-w[x][y];
        if(tmp==0){
            visy[y]=1;
            if(linker[y]==-1 || DFS(linker[y])){
                linker[y]=x;
                return 1;
            }
        }else if(slack[y]>tmp){
            slack[y]=tmp;
        }
    }
    return 0;
}

int KM(){
    int i,j;
    memset(linker,-1,sizeof(linker));
    memset(ly,0,sizeof(ly));
    for(i=1;i<=nx;i++)
        for(j=1,lx[i]=-INF;j<=ny;j++)
            if(w[i][j]>lx[i])
                lx[i]=w[i][j];
    for(int x=1;x<=nx;x++){
        for(i=1;i<=ny;i++)
            slack[i]=INF;
        while(1){
            memset(visx,0,sizeof(visx));
            memset(visy,0,sizeof(visy));
            if(DFS(x))
                break;
            int d=INF;
            for(i=1;i<=ny;i++)
                if(!visy[i] && d>slack[i])
                    d=slack[i];
            for(i=1;i<=ny;i++)
                if(visx[i])
                    lx[i]-=d;
            for(i=1;i<=ny;i++)
                if(visy[i])
                    ly[i]+=d;
                else
                    slack[i]-=d;
        }
    }
    int res=0;
    for(i=1;i<=ny;i++)
        if(linker[i]!=-1)
            res+=w[linker[i]][i];
    return -res;
}

int main(){

    //freopen("input.txt","r",stdin);

    int t;
    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&n,&m);
        nx=ny=n;
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++)
                w[i][j]=-INF;   //即将边的权值取反
        int u,v,cap;
        while(m--){
            scanf("%d%d%d",&u,&v,&cap);
            if(w[u][v]<-cap)
                w[u][v]=-cap;
        }
        printf("%d\n",KM());
    }
    return 0;
}