hdu 1853

Cyclic Tour

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/65535 K (Java/Others)
Total Submission(s): 861    Accepted Submission(s): 438

Problem Description

There are N cities in our country, and M one-way roads connecting them. Now Little Tom wants to make several cyclic tours, which satisfy that, each cycle contain at least two cities, and each city belongs to one cycle exactly. Tom wants the total length of all the tours minimum, but he is too lazy to calculate. Can you help him?

 

Input

There are several test cases in the input. You should process to the end of file (EOF).
The first line of each test case contains two integers N (N ≤ 100) and M, indicating the number of cities and the number of roads. The M lines followed, each of them contains three numbers A, B, and C, indicating that there is a road from city A to city B, whose length is C. (1 ≤ A,B ≤ N, A ≠ B, 1 ≤ C ≤ 1000).

 

Output

Output one number for each test case, indicating the minimum length of all the tours. If there are no such tours, output -1.

 

Sample Input

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 6 5 1 2 1 2 3 1 3 4 1 4 5 1 5 6 1

 

Sample Output

42 -1

Hint

In the first sample, there are two cycles, (1->2->3->1) and (6->5->4->6) whose length is 20 + 22 = 42.

類似hdu3488，只是加了一個判斷是否都被匹配。

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define INF 9999999
#define maxn 205
int lx[maxn],ly[maxn];
int w[maxn][maxn];
int Match[maxn],visx[maxn],visy[maxn];
int slack[maxn];
int ans,n,m;
bool findPath(int x)
{
    int tmp;
    visx[x]=1;
    for(int y=1; y<=n; y++)
    {
        if(visy[y])continue;
        if(w[x][y]==lx[x]+ly[y])
        {
            visy[y]=1;
            if(!Match[y]||findPath(Match[y]))
            {
                Match[y]=x;
                return true;
            }
        }
        else
            slack[y]=min(slack[y],w[x][y]-lx[x]-ly[y]);
    }
    return false;
}
void km()
{
    memset(Match,0,sizeof(Match));
    memset(ly,0,sizeof(ly));
    for(int i=1; i<=n; i++)
    {
        lx[i]=INF;
        for(int j=1; j<=n; j++)
            lx[i]=min(lx[i],w[i][j]);
    }
    for(int x=1; x<=n; x++)
    {
        for(int i=1; i<=n; i++)
            slack[i]=INF;
        while(true)
        {
            memset(visx,0,sizeof(visx));
            memset(visy,0,sizeof(visy));
            if(findPath(x))break;
            int tmp=INF;
            for(int i=1; i<=n; i++)
            {
                if(!visy[i])
                {
                    if(tmp>slack[i])
                        tmp=slack[i];
                }
            }
            if(tmp==INF)return ;
            for(int i=1; i<=n; i++)
            {
                if(visx[i])lx[i]+=tmp;
                if(visy[i])ly[i]-=tmp;
            }
        }
    }
}
int main()
{
    int cas;
    while(scanf("%d %d",&n,&m)!=EOF)
    {
        int a,b,c;

        for(int i=1; i<=n; i++)
            for(int j=1; j<=n; j++)
            {
                w[i][j]=INF;
            }
        for(int i=0; i<m; i++)
        {
            cin>>a>>b>>c;
            if(w[a][b]>c)
                w[a][b]=c;
        }
        km();
        int ans=0;
        int flag=0;
        for(int i=1; i<=n; i++)
        {
            if(Match[i]==0||w[Match[i]][i]==INF)
            {
                flag=1;
                cout<<"-1"<<endl;
                break;
            }
            if(Match[i])
                ans+=w[Match[i]][i];
        }
        if(!flag)
        cout<<ans<<endl;
    }
    return 0;
}