HDOJ 5418 Victor and World 状压DP

水状压DP

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/131072 K (Java/Others)
Total Submission(s): 407    Accepted Submission(s): 164

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  countries on the earth, which are numbered from . They are connected by  undirected flights, detailedly the -th flight connects the -th and the -th country, and it will cost Victor's airplane  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 , 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 , denoting the number of test cases.
In every test case, there are two integers  in the first line, denoting the number of the countries and the number of the flights.

Then there are  lines, each line contains three integers , describing a flight.

.

 

Output

Your program should print  lines : the -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)

 

/* ***********************************************
Author        :CKboss
Created Time  :2015年08月23日 星期日 10时27分21秒
File Name     :HDOJ5418.cpp
************************************************ */

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <cassert>

using namespace std;

typedef long long int LL;

const int INF=0x3f3f3f3f;

int n,m;
int G[18][18];
int dp[18][1<<18];

int main()
{
	//freopen("in.txt","r",stdin);
	//freopen("out.txt","w",stdout);

	int T_T;
	scanf("%d",&T_T);
	while(T_T--)
	{
		scanf("%d%d",&n,&m);
		memset(G,63,sizeof(G));
		for(int i=0;i<n;i++) G[i][i]=0;
		for(int i=0,a,b,c;i<m;i++)
		{
			scanf("%d%d%d",&a,&b,&c);
			a--; b--;
			G[a][b]=min(G[a][b],c);
			G[b][a]=min(G[b][a],c);
		}

		for(int k=0;k<n;k++)
			for(int i=0;i<n;i++)
				for(int j=0;j<n;j++)
					G[i][j]=min(G[i][j],G[i][k]+G[k][j]);
		memset(dp,63,sizeof(dp));

		dp[0][1]=0;
		for(int i=1;i<(1<<n);i++) /// status
		{
			for(int j=0;j<n;j++) /// from point
			{
				if(((1<<j)&i)!=0)
				{
					for(int k=0;k<n;k++) /// to point
					{
						if(((1<<k)&i)==0)
						{
							dp[k][((1<<k)|i)]=min(dp[k][((1<<k)|i)],dp[j][i]+G[j][k]);
						}
					}
				}
			}
		}
		int ans=INF;
		for(int i=0;i<n;i++)
		{
			ans=min(ans,dp[i][(1<<n)-1]+G[i][0]);
		}
		if(n==1) ans=0;
		printf("%d\n",ans);
	}
    return 0;
}