HDU 3001 traveling

最新推荐文章于 2018-09-25 20:20:09 发布

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本文探讨了一个特殊的旅行商问题，目标是在给定的城市和路径费用中找到一条路径，使得每个城市最多被访问两次，并且总费用最小。通过使用状态压缩和动态规划的方法解决了这一问题。

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Travelling

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8481 Accepted Submission(s): 2759

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

Sample Output

100
90
7

题意：n 个点， m 条边，走遍 n 个点，每个点最多走2次，求最小代价。n <= 10

每个点走两次这个条件貌似很难用，其实转化成3进制数就可以像普通tsp问题那样做了！和普通状压的区别就是多了个预处理，也不能直接用位运算而已

#include <cstdio>
#include <string.h>
#define min(a, b) ((a) < (b) ? (a) : (b))
int f[15][60005], three[14], dig[60005][15];
int mp[105][105];
void pre(){
	int i, j, s = 1, t;
	three[0] = 1;
	for(i = 1; i <= 11; i++) three[i] = three[i-1] * 3;
	for(i = 1; i < 59049; i++)
		for(t = i, j = 0; j < 10; j++) dig[i][j] = t % 3, t /= 3;
}
int main(){
	int i, j, n, m, a, b, c, ans, k;
	pre(); 
	while(scanf("%d%d", &n, &m) != EOF){
		ans = 2e9;
		memset(mp, 0, sizeof(mp));
		memset(f, 120, sizeof(f));
		for(i = 1; i <= n; i++) f[i][three[i-1]] = 0;
		for(i = 1; i <= m; i++){
			scanf("%d%d%d", &a, &b, &c);
			if(!mp[a][b]) mp[a][b] = mp[b][a] = c;
			else if(mp[a][b] > c) mp[a][b] = mp[b][a] = c;
		}
		for(i = 1; i < three[n]; i++)
			for(j = 0; j < n; j++) if(dig[i][j] && f[j+1][i] < 2e9)
				for(k = 0; k < n; k++)
					if(mp[j+1][k+1] && dig[i][k] < 2) f[k+1][i+three[k]] = min(f[k+1][i+three[k]], f[j+1][i] + mp[j+1][k+1]);
		for(i = 1; i < three[n]; i++){
			for(j = 0; j < n; j++) if(!dig[i][j]) break;
			if(j == n){
				for(j = 0; j < n; j++) ans = min(ans, f[j][i]);
			}
		}
		if(ans < 2e9) printf("%d\n", ans);
		else printf("-1\n");
	}
}