HDU 5418 Victor and World（状态压缩路径DP或+floyd 两种方法 求 玈行商问题）经典

Victor and World

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/131072 K (Java/Others)
Total Submission(s): 174    Accepted Submission(s): 79

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) 

题意：有n个点m条无向边，现在从1点出发最后又回到1点，必须经过所有的点。

方法一：状态压缩DP +floyd。（500+MS）AC

#include
#include
const int INF = 1<<30 ;

int dp[1<<17][20];
int n,m,u,v,w,mp[20][20] ;
void floyd()
{
    for(int e=0; emp[i][e]+mp[e][j])
              mp[i][j] = mp[i][e]+mp[e][j];
    }
}
int main()
{

    int T ;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        for(int i=0; i<(1<w)
                mp[u][v]=mp[v][u]=w;
        }
        floyd();
        dp[1][0] = 0 ;
        for(int s=1; s<(1<dp[s][i]+mp[i][j])
                    dp[s|(1< dp[(1<

方法二：状态压缩DP  （900+MS）AC

#include
#include
const int INF = 1<<30 ;

int dp[1<<17][20];
int n,m,u,v,w,mp[20][20] , kk[1<<17];

int main()
{

    int T ;
    for(int i=0; i<(1<<16); i++)
    {
        kk[i]=0;
        for(int j=0; j<=16; j++)
            if(i&(1<w)
                mp[u][v]=mp[v][u]=w;
        }
        dp[1][0] = 0 ;
        for(int s=1; s<(1<dp[s][i]+mp[i][j])
                    dp[s|(1<