hdu 5418 Victor and World 状态压缩dp，旅行商问题_after trying hard for many years, victor has final-优快云博客

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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

Sample Output

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;
}




/*
33
4
10
1 2 1
1 3 1
1 4 10
2 3 1
2 4 10
3 4 1
1 3 10
1 4 10
1 2 10
1 3 10


3 2
1 2 2
1 3 3
*/