HDU 3001 Travelling

After coding so many days,Mr Acmer wants to have a good rest.So travelling is the best choice!He has decided to visit n cities(he insists on seeing all the cities!And he does not mind which city being his start station because superman can bring him to any city at first but only once.), and of course there are m roads here,following a fee as usual.But Mr Acmer gets bored so easily that he doesn't want to visit a city more than twice!And he is so mean that he wants to minimize the total fee!He is lazy you see.So he turns to you for help. Input

There are several test cases,the first line is two intergers n(1<=n<=10) and m,which means he needs to visit n cities and there are m roads he can choose,then m lines follow,each line will include three intergers a,b and c(1<=a,b<=n),means there is a road between a and b and the cost is of course c.Input to the End Of File.

Output

Output the minimum fee that he should pay,or -1 if he can't find such a route.

Sample Input

2 1
1 2 100
3 2
1 2 40
2 3 50
3 3
1 2 3
1 3 4
2 3 10

Sample Output

100
90
7 

  因为每个点最多走过两次，所以用三进制进行状态压缩。

#include<iostream>//三进制状态压缩DP
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int bit[11]={1,3,9,27,81,243,729,2187,6561,19683,59049};//一共3的10次方个状态
int dp[60000][12];//第一维代表状态，第二维代表当前位置
int num[60000][12];
int map[12][12];
const int INF=0x3f3f3f3f;
int n,m;
int nmin;
void Three()
{
    int i,j,b;
    for(i=0;i<=bit[10]-1;i++)
    {
        b=i;
        for(j=0;j<=9;j++)
        {
            num[i][j]=b%3;
            b/=3;
        }
    }
}
void Init()
{
    int i,j;
    memset(map,-1,sizeof(map));
    for(i=0;i<=bit[n]-1;i++)
    {
        for(j=0;j<=n-1;j++)
        {
            dp[i][j]=INF;
        }
    }
    for(j=0;j<=n-1;j++)
    {
        dp[bit[j]][j]=0;
    }
}
void DP()
{
    int i,j,flag,k,next;
    nmin=INF;
    for(i=0;i<=bit[n]-1;i++)
    {
        flag=1;
        for(j=0;j<=n-1;j++)
        {
            if(num[i][j]==0)
                flag=0;
            if(dp[i][j]==INF)
                continue;
            for(k=0;k<=n-1;k++)
            {
                if(j==k||num[i][k]>=2||map[k][j]==-1)
                    continue;
                next=i+bit[k];
                dp[next][k]=min(dp[next][k],dp[i][j]+map[j][k]);
            }
        }
        if(flag)
        {
            for(j=0;j<=n-1;j++)
            {
                nmin=min(nmin,dp[i][j]);//该状态下的最优解
            }
        }
    }
    return ;
}
int main()
{
    int a,b,c;
    Three();
    while(~scanf("%d%d",&n,&m))
    {
        Init();
        int i,j;
        for(i=0;i<=m-1;i++)
        {
            scanf("%d%d%d",&a,&b,&c);
            if(map[a-1][b-1]==-1)
                map[a-1][b-1]=map[b-1][a-1]=c;
            else map[a-1][b-1]=map[b-1][a-1]=min(map[a-1][b-1],c);
        }
        DP();
        if(nmin==INF) nmin=-1;
        printf("%d\n",nmin);
    }
    return 0;
}