hdu 5418 Victor and World 狀態壓縮dp，旅行商問題

Victor and World Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/131072 K (Java/Others)
Total Submission(s): 385    Accepted Submission(s): 160


Problem Description After trying hard for many years, Victor has finally received a pilot license. To have a celebration, he intends to buy himself an airplane and fly around the world. There are  n countries on the earth, which are numbered from  1 to  n. They are connected by  m undirected flights, detailedly the  i-th flight connects the  ui-th and the  vi-th country, and it will cost Victor's airplane  wi L fuel if Victor flies through it. And it is possible for him to fly to every country from the first country.

Victor now is at the country whose number is  1, he wants to know the minimal amount of fuel for him to visit every country at least once and finally return to the first country.  
Input The first line of the input contains an integer  T, denoting the number of test cases.
In every test case, there are two integers  n and  m in the first line, denoting the number of the countries and the number of the flights.

Then there are  m lines, each line contains three integers  ui,  vi and  wi, describing a flight.

1≤T≤20.

1≤n≤16.

1≤m≤100000.

1≤wi≤100.

1≤ui,vi≤n.  
Output Your program should print  T lines : the  i-th of these should contain a single integer, denoting the minimal amount of fuel for Victor to finish the travel.  
Sample Input

1 3 2 1 2 2 1 3 3  
Sample Output

10  
Source BestCoder Round #52 (div.2)

#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>using namespace std;int map[20][20];int dp[1<<17][20];int main(){    int t,n,m,w,u,v;    scanf("%d",&t);    while(t--){        scanf("%d%d",&n,&m);        memset(map,0x3f,sizeof(map));        for(int i = 0;i < n; i++)            map[i][i] = 0;        for(int i = 0;i < m; i++){            scanf("%d%d%d",&u,&v,&w);            u--,v--;            map[u][v] = map[v][u] = min(map[u][v],w);        }        for(int i = 0;i < n;i++)            for(int j = 0;j < n; j++)                for(int k = 0;k < n; k++)                    map[j][k] = map[k][j] = min(map[j][k],map[j][i]+map[i][k]);        memset(dp,0x3f,sizeof(dp));        dp[1][0] = 0;        for(int i = 1;i < (1<<n);i++){            for(int j = 0;j < n; j++){                if(i&(1<<j)){                    for(int k = 0;k < n; k++){                        if((i&(1<<k))==0 && j != k){                           dp[i|(1<<k)][k] = min(dp[i|(1<<k)][k],dp[i][j]+map[j][k]);                        }                    }                }            }        }        int ans = dp[0][0];        for(int i = 0;i < n; i++){            ans = min(ans,dp[(1<<n)-1][i]+map[i][0]);        }        printf("%d\n",ans);    }    return 0;}/*334101 2 11 3 11 4 102 3 12 4 103 4 11 3 101 4 101 2 101 3 103 21 2 21 3 3*/