hdu 1853 Cyclic Tour //km

Cyclic Tour

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/65535 K (Java/Others)
Total Submission(s): 93    Accepted Submission(s): 51

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. 
    
    
    
    
     
   
   
   
   

 

Author

RoBa@TJU

 

Source

HDU 2007 Programming Contest - Final

 

Recommend

lcy

 

 

 

#include<cstdio>

#include<cstring>

#include<algorithm>

const int inf=1<<28;

using namespace std;

int g[505][505],lx[505],ly[505];//顶点标号

bool sx[505],sy[505];//是否已经搜索过

int link[505],n,stack[505];

bool path(int k)//从x[k]寻找增广路

{

    sx[k]=true;

    for(int i=1; i<=n; i++)

    {

        if(sy[i])  continue;

        int t=ly[i]+lx[k]-g[k][i];

        if(t==0)

        {

            sy[i]=1;

            if(link[i]==-1||path(link[i]))

            {

                link[i]=k;

                return true;

            }

        }

        else if(stack[i]>t) stack[i]=t;

    }

    return false;

}

int BestMatch()

{

    int d,sum;

    memset(ly,0,sizeof(ly));

    memset(link,-1,sizeof(link));

    for(int i=1; i<=n; i++)

    {

        lx[i]=-inf;

        for(int j=1; j<=n; j++)

            if(lx[i]<g[i][j]&&g[i][j]!=0)  lx[i]=g[i][j];

    }

    for(int k=1; k<=n; k++)

    {

        for (int i=1; i<=n; i++) stack[i]=inf;

        while(1)

        {

            memset(sx,0,sizeof(sx));

            memset(sy,0,sizeof(sy));

            if(path(k))  break;

            d=inf;

            for(int i=1; i<=n; i++)

                if (!sy[i]&&stack[i]<d) d=stack[i];

            for(int i=1; i<=n; i++)

                if(sx[i]) lx[i]-=d;

            for(int i=1; i<=n; i++)

                if(sy[i]) ly[i]+=d;

                else  stack[i]-=d;

        }

    }

    sum=0;

    for(int i=1; i<=n; i++)

    {

        if(link[i]==-1||g[link[i]][i]==-inf)  return -1;

        sum+=g[link[i]][i];

    }

    return -sum;

}

int main()

{

    int m;

    while(scanf("%d%d",&n,&m)!=EOF)

    {

        for(int i=1;i<=n;i++)

          for(int j=1;j<=n;j++)

                  g[i][j]=inf;

        for(int i=1; i<=m; i++)

        {

            int x,y,c;

            scanf("%d%d%d",&x,&y,&c);

            g[x][y]=min(g[x][y],c);

        }

        for(int i=1; i<=n; i++)

            for(int j=1; j<=n; j++) g[i][j]=-g[i][j];

        printf("%d/n",BestMatch());

    }

    return 0;

}