HDU 3001 Travelling（状态压缩DP）

思路：类似TSP问题，不过每个点可以经过不超过两次，那么就可以考虑用三进制来压缩状态，先预处理出每个数的三进制的第j位是什么，然后其他就和TSP类似的

坑点：有重边

#include<bits/stdc++.h>
using namespace std;
const int maxn = 60000;
#define inf 0x3f3f3f
int three[]={0,1,3,9,27,81,243,729,2187,6561,19683,59049};
int n,m;
int mp[12][12];
int dp[maxn][12];
int dig[maxn][12];   //3进制数i的第j位是什么
void init()
{
    for(int i = 0;i<maxn;i++)
	{
		int t = i;
		for(int j = 1;j<=10;j++)
		{
			dig[i][j]=t%3;
			t/=3;
			if(!t)break;
		}
	}
}

int main()
{
	init();
    while(scanf("%d%d",&n,&m)!=EOF)
	{
		memset(mp,0,sizeof(mp));
		for(int i= 1;i<=m;i++)
		{
			int u,v,w;
			scanf("%d%d%d",&u,&v,&w);
			if(mp[u][v]==0)
			{
				mp[u][v]=w;
				mp[v][u]=w;
			}
			if(w<mp[u][v] && mp[u][v]!=0)
			{
				mp[u][v]=w;
		 	    mp[v][u]=w;
			}
		}
		int ans = inf;
		memset(dp,inf,sizeof(dp));
		for(int i = 1;i<=n;i++)
			dp[three[i]][i]=0;
		for(int s = 1;s<three[n+1];s++)
		{
			int flag = 1;
            for(int i = 1;i<=n;i++)
			{
				if(!dig[s][i])flag=0;
				if(dp[s][i]==inf)continue;
				for(int j = 1;j<=n;j++)
                {
					if(i==j)continue;
					if(dig[s][j]>=2 || !mp[i][j])continue;
					dp[s+three[j]][j]=min(dp[s][i]+mp[i][j],dp[s+three[j]][j]);
				}
			}
			if(flag)
				for(int i = 1;i<=n;i++)
				ans = min(ans,dp[s][i]);
		}
		if(ans==inf)
			printf("-1\n");
		else
			printf("%d\n",ans);
	}
}

Description

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