HDU 3001Travelling （状压DP）



Travelling

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3686    Accepted Submission(s): 1151

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 

  

 思路：TSP类问题，注意每个点最多走两次，所以要用三进制表示。可以预处理出状态的每一位是什么

#include <iomanip>#include <cstdio>#include <map>#include <queue>#include <cstring>#include <iostream>#include <vector>#include <cmath>#include <string>using namespace std;#define LL long long const int mod =1e8;#define inf 0x1f1f1f1f#define lson id<<1,l,mid  #define rson id<<1|1,mid+1,rint fin[12]={0,1,3,9,27,81,243,729,2187,6561,19683,59049};int getdigit[59049+10][11],dp[59049+10][11],dis[11][11],n,m,vis[11][11];int main(){for(int s=0;s<59050;s++){int temp=s;for(int i=1;i<=10;i++){getdigit[s][i]=temp%3;temp/=3;if(temp==0) break;}}while(scanf("%d%d",&n,&m)!=EOF){int a,b,c;memset(vis,0,sizeof(vis));memset(dp,inf,sizeof(dp));memset(dis,inf,sizeof(dis));for(int i=1;i<=m;i++){scanf("%d%d%d",&a,&b,&c);if(vis[a][b]){dis[a][b]=min(dis[a][b],c);dis[b][a]=dis[a][b];}else {dis[a][b]=dis[b][a]=c;vis[a][b]=vis[b][a]=1;}}bool flag;int ans=inf;for(int i=1;i<=n;i++) dp[fin[i]][i]=0;for(int s=0;s<fin[n+1];s++){flag=1;for(int i=1;i<=n;i++){if(getdigit[s][i]==0) flag=0;if(dp[s][i]==inf) continue;for(int j=1;j<=n;j++){if(j==i) continue;if(dis[i][j]==inf||getdigit[s][j]>=2) continue;dp[s+(fin[j])][j]=min(dp[s+(fin[j])][j],dp[s][i]+dis[i][j]);}}if(flag){for(int i=1;i<=n;i++){ans=min(ans,dp[s][i]);}}}if(ans==inf) puts("-1");else printf("%d\n",ans);}return 0;}