ACM: HDU 5418 Victor and World

HDU 5418 Victor and World

Time Limit:2000MS     Memory Limit:131072KB     64bit IO Format:%I64d & %I64u

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  to . 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  and  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 ,  and , 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

/*/
一开始题目没读仔细，以为是一个最小树，秒WA一发；
后来想半天发现这个有环，就不是最小树了，搜了一下是Floyd+dp状压。

写完之后一直发现输出的是INF=0x3f3f3f3f ，找了半天最后想到二进制标记状态这里，maps标记用0 0开始会舒服好多。改过来就对了。题目很有意思。。

AC代码：
/*/

#include"algorithm"
#include"iostream"
#include"cstring"
#include"cstdlib"
#include"string"
#include"cstdio"
#include"vector"
#include"cmath"
#include"queue"
using namespace std;
typedef long long LL;
#define memset(x,y) memset(x,y,sizeof(x))
#define memcpy(x,y) memcpy(x,y,sizeof(x))
#define MX 401
#define INF 0x3f3f3f3f

int maps[20][20];
int dp[200000][20],vis[20];

void init() {

	memset(maps,0x3f);
	memset(vis,0x3f);
	memset(dp ,0x3f);
}

int main() {
	int T;
	scanf("%d",&T);
	while(T--) {
		init();
		int n,m,u,v,w;
		scanf("%d%d",&n,&m);
		for(int i=0; i<m; i++) {
			scanf("%d%d%d",&u,&v,&w);
			if(maps[--u][--v] > w)    //状态压缩从0开始会好写一些 
				maps[v][u]=maps[u][v]= w;
		}
		for(int i=0; i<n; i++)maps[i][i]=0; //标记自己到自己距离为0   //后面会要加到这个数字。。
		for(int k=0; k<n; k++) {
			for(int i=0; i<n; i++) {
				for(int j=0; j<n; j++) {
					maps[i][j]=min(maps[i][j],maps[i][k]+maps[k][j]); //Floyd 把去某一点的最小路程计算出来
				}
			}
		}
		dp[1][0]=0;
		vis[0]=0;
		m=1<<n;
		for(int i=1; i<m; i++) {
			for(int j=0; j<n; j++) {
				if(dp[i][j]==INF)continue;
				for(int k=0; k<n; k++) {
					if(i&(1<<k)||maps[j][k]==INF)continue;  //二进制 1 表示该点走过，0表示没走过，第二维表示现在所在的点。压缩状态
					if( dp[i|(1<<k)][k]>dp[i][j]+maps[j][k]) {
						dp[i|(1<<k)][k]=dp[i][j]+maps[j][k];
						vis[k]=min(vis[k],dp[i|(1<<k)][k]); //比较，去到下一点需要的最小路
					}
				}
			}
		}
		int minn=1e9+100;;
		for(int i=0; i<n; i++) {
			minn=min(minn,dp[m-1][i]+vis[i]);
		}
		printf("%d
",minn);
	}
	return 0;
}