hdu3001——Travelling 三进制TSP, 状态压缩

Travelling

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4106    Accepted Submission(s): 1310

Problem Description

After coding so many days,Mr Acmer wants to have a good rest.So travelling is the best choice!He has decided to visit n cities(he insists on seeing all the cities!And he does not mind which city being his start station because superman can bring him to any city at first but only once.), and of course there are m roads here,following a fee as usual.But Mr Acmer gets bored so easily that he doesn't want to visit a city more than twice!And he is so mean that he wants to minimize the total fee!He is lazy you see.So he turns to you for help.

 

Input

There are several test cases,the first line is two intergers n(1<=n<=10) and m,which means he needs to visit n cities and there are m roads he can choose,then m lines follow,each line will include three intergers a,b and c(1<=a,b<=n),means there is a road between a and b and the cost is of course c.Input to the End Of File.

 

Output

Output the minimum fee that he should pay,or -1 if he can't find such a route.

 

Sample Input

   
   
   
   

    
    
    
    2 1
1 2 100
3 2
1 2 40
2 3 50
3 3
1 2 3
1 3 4
2 3 10
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    100
90
7 
   
   
   
   

 

Source

2009 Multi-University Training Contest 11 - Host by HRBEU

 

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这题一看就是TSP，然后想到要状压，但是这里每个城市最多可以走两次，所以不能用普通的二进制来状压，需要三进制

三进制每一位是0 1 2， 0表示这座城市没到过，1表示到过一次，2表示到过两次
由于一共10座城市，我们先把3^0 --- 3 ^ 10预处理出来，然后化作三进制存好，接着就是dp，但是要注意，可以作为最后状态的状态，它的每一位都不为0（否则就有某个城市没到过），最后枚举所有可以作为尾状态的状态然后求个最小值就ok了

#include <map>
#include <set>
#include <list>
#include <stack>
#include <queue>
#include <vector>
#include <cmath>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int inf = 0x3f3f3f3f;

int dp[60000][15];
int dist[15][15];
int three[15] = {0, 1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049};
int city[60000][15];

int main()
{
	int n, m, u, v, w;
	memset (city, inf, sizeof(city));
	for (int i = 0; i < 59059; ++i)
	{
		int tmp = i;
		for (int j = 1; j <= 10; ++j)
		{
			city[i][j] = tmp % 3;
			tmp /= 3;
		}
	}
	while (~scanf("%d%d", &n, &m))
	{
		memset (dist, inf, sizeof(dist));
		memset (dp, inf, sizeof(dp));
		for (int i = 1; i <= n; ++i)
		{
			dp[three[i]][i] = 0;
		}
		dp[0][0] = 0;
		for (int i = 1; i <= m; ++i)
		{
			scanf("%d%d%d", &u, &v, &w);
			dist[u][v] = min(dist[u][v], w);
			dist[v][u] = dist[u][v];
		}
		int ans = inf;
		for (int i = 0; i < three[n + 1]; ++i)
		{
			bool is_end = 1;
			for (int j = 1; j <= n; ++j)
			{
				if (city[i][j] == 0)
				{
					is_end = 0;
				}
				for (int k = 1; k <= 10; ++k)
				{
					if (city[i][j] == 2)
					{
						continue;
					}
					dp[i + three[j]][j] = min(dp[i + three[j]][j], dp[i][k] + dist[k][j]);
				}
			}
			if (is_end)
			{
				for (int j = 1; j <= n; ++j)
				{
					ans = min(ans, dp[i][j]);
				}
			}
		}
		if (ans == inf)
		{
			printf("-1\n");
			continue;
		}
		printf("%d\n", ans);
	}
	return 0;
}