hdu 5418 Victor and World 最短路 floyd 解题报告

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

13 21 2 21 3 3

 

Sample Output

10

 

Source

BestCoder Round #52 (div.2)

 

思路：http://blog.csdn.net/qq_21899803/article/details/47903197

代码：

#include<iostream>#include<cstdio>#include<cstring>using namespace std;#define maxn 17#define inf 0x3f3f3f3fint Map[maxn][maxn],dp[maxn][1<<maxn|1],n;void floyd(){    for(int k=1;k<=n;++k)        for(int i=1;i<=n;++i)            for(int j=1;j<=n;++j)                if(Map[i][j]>Map[i][k]+Map[k][j])                    Map[i][j]=Map[i][k]+Map[k][j];}int main(){    int T,x,y,z,i,j,k,m;    cin>>T;    while(T--)    {        scanf("%d%d",&n,&m);        memset(Map,inf,sizeof(Map));        for(i=0;i<m;++i){            scanf("%d%d%d",&x,&y,&z);            if(z<Map[x][y]) Map[x][y]=Map[y][x]=z;        }        floyd();        memset(dp,inf,sizeof(dp));        dp[1][1]=0;        for(i=1;i<(1<<n);++i)   //依次枚举各种状态            for(j=1;j<=n;++j)            {                if(i&(1<<(j-1)))   //枚举该状态下访问的顶点                {                    for(k=1;k<=n;++k)   //枚举访问的其它顶点                        if(i&(1<<(k-1))&&j!=k)                            dp[j][i]=min(dp[j][i],dp[k][i^(1<<(j-1))]+Map[k][j]);                }            }        int ans=inf;        for(i=2;i<=n;++i) ans=min(ans,dp[i][(1<<n)-1]+Map[i][1]);        printf("%d\n",n==1?0:ans);    }    return 0;}