hdoj3001Travelling【状压dp 类TSP】

 

Travelling

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5171    Accepted Submission(s): 1691

Problem 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 11 2 1003 21 2 402 3 503 31 2 31 3 42 3 10

 

Sample Output

100907 

 

题意：类TSP问题每个点最多走两次

#include<cstdio>#include<cstdlib>#include<cstring>#include<cmath>#define inf 0x3f3f3f3fusing namespace std;int MIN(int a,int b){return a<b?a:b;}int state[15];int map[15][15];int dp[60000][15]; int dig[60000][15];void init(){state[1]=1;for(int i=2;i<12;++i){state[i]=state[i-1]*3;}for(int i=1;i<=state[11];++i){//将该状态变换为三进制; int n=i,k=1;while(n){dig[i][k++]=n%3;n/=3;}}}int main(){init();int i,j,k,n,m;while(scanf("%d%d",&n,&m)!=EOF){memset(dp,0x3f,sizeof(dp));memset(map,0x3f,sizeof(map));for(i=1;i<=m;++i){int a,b,c;scanf("%d%d%d",&a,&b,&c);if(map[a][b]>c){map[a][b]=map[b][a]=c;}}for(i=1;i<=n;++i)dp[state[i]][i]=0;//确定起点 int ans=inf;for(int s=state[1];s<state[n+1];++s){//枚举各种状态; int flag=1;for(i=1;i<=n;++i){if(dig[s][i]==0)flag=0;//判断是否经过该点 if(dp[s][i]==inf)continue;for(j=1;j<=n;++j){if(i==j)continue;if(map[i][j]==inf||dig[s][j]==2)continue;//如若不能通行或经过该地超过两次 int news=s+state[j];//更新 dp[news][j]=MIN(dp[news][j],dp[s][i]+map[i][j]);}}if(flag){//如果该状态下所有点都经过过更新最小花费; for(j=1;j<=n;++j){ans=MIN(ans,dp[s][j]);}}}if(ans==inf)printf("-1\n");else printf("%d\n",ans);}return 0;}