hdu3001 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.

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 the minimum fee that he should pay,or -1 if he can‘t find such a route.

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

100
90
7 

看了别人的思路，自己写出来了：）。这题和poj3311差不多，但是不能用floyd处理，因为它有访问次数限制，最多相同的地方访问两次，至少一次，所以为了存储状态，我们可以用三进制表示，1代表访问一次，2代表访问2次。动态转移方程也和之前的差不多，为dp[s][i]=min(dp[s][i],dp[s-san[i-1]][j]+dis[j][i])。最后的结论要在访问过程中产生，如果当前所枚举的状态符合用三进制表示后每一位的数都大于0，那么就表示都访问到了，就可以和所求的结果ans比，如果比ans小就更新。
#include<stdio.h>
#include<string.h>
#define inf 88888888
int dis[15][15],dp[200000][15],wei[15],num,t;
int san[15]={1,3,9,27,81,243,729,2187,6561,19683,59049,177147};
int min(int a,int b){
	return a<b?a:b;
}
void zhuanhua(int x)
{
	int i,j;
	num=0,t=0;
	while(x>0){
		if(x%3>0)num++;
		wei[++t]=x%3;
		x=x/3;
	}
}

int main()
{
	int n,m,i,j,a,b,c,s,ans;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		for(i=1;i<=n;i++){
			for(j=1;j<=n;j++){
				dis[i][j]=inf;
			}
		}
		for(i=1;i<=m;i++){
			scanf("%d%d%d",&a,&b,&c);
		    if(dis[a][b]>c)dis[a][b]=dis[b][a]=c;
		}
		ans=inf;
		for(s=1;s<san[n+1];s++){
			for(i=1;i<=n;i++){
				dp[s][i]=inf;
			}
		}
		
		
		for(s=1;s<san[n+1];s++){
			zhuanhua(s);
			for(i=1;i<=n;i++){
				if(t>=i && wei[t]>0){
					if(s==san[i-1]){
						dp[s][i]=0;
					}
					else{
						for(j=1;j<=n;j++){
							if(j!=i && wei[j]>0){
								dp[s][i]=min(dp[s][i],dp[s-san[i-1]][j]+dis[j][i]);
								if(num==n){
									ans=min(ans,dp[s][i]);
								}
							}
						}
					}
				}
			}
			if(ans==0)break;
		}
		if(ans==inf)printf("-1\n");
		else printf("%d\n",ans);
	}
}