hdu3001 三进制状态压缩+dp

最新推荐文章于 2020-04-05 20:01:56 发布

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#状态压缩+dp

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本文介绍了一个旅行路径规划问题，目标是最小化访问多个城市总费用的同时确保每个城市最多被访问两次。采用三进制状态表示法来跟踪每个城市的访问次数，并通过动态规划求解最优路径。

Travelling

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3477 Accepted Submission(s): 1084

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

单排(dp)果然博大精深。三进制状态的值表示当前点已取第几次。哦，先说下题意吧。。。题意应该就是有个妹纸去旅游，然后有N个地方，她要游遍所有的地方但是每个地方最多只能去两次！问最短距离是多少。两次，第一次写被这两次坑了，没注意到，后来三进制可以表示到一个地方去了几次，当所有地方都遍历到，就可以进行取值，当然要是最小的。条件也有几个。初始点要设为0.

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cmath>
#include <queue>
#include <map>
#include <stack>
#include <list>
#include <vector>
using namespace std;
#define LL __int64
int dis[15][15],kfc[60000][15],dp[60000][15];
int t,n,j,m,i,u,v,d;
int sta[12]={0,1,3,9,27,81,243,729,2187,6561,19683,59049};
#define INF 0x7fffff1f
#define min(a,b) a>b?b:a
int main()
{
	for (i=0;i<59049;i++)
	{
		t=i;
		for (j=1;j<=10;j++)
		{
			kfc[i][j]=t % 3;
			t/=3;
			if (t==0) break;
		}
	}
	while (2==scanf("%d%d",&n,&m))
	{
		for (i=1;i<60000;i++)
			for (j=1;j<=12;j++)
			{
				if (i<15)
					dis[i][j]=INF;
				dp[i][j]=INF;
			}
		for (i=1;i<=m;i++)
		{
			scanf("%d%d%d",&u,&v,&d);
			dis[u][v]=dis[v][u]=min(dis[u][v],d);
		}
		for (i=1;i<=n;i++)
			dp[sta[i]][i]=0;
		int ans=INF;	
		for (t=0;t<sta[n+1];t++)
		{
			int find=1;
			for (i=1;i<=n;i++)
			{
				if (kfc[t][i]==0) find=0;
				if (dp[t][i]==INF) continue;
				for (j=1;j<=n;j++)
				{
					if (j==i || kfc[t][j]==2 || dis[i][j]==INF) continue;
					int newt=t+sta[j];
					dp[newt][j]=min(dp[newt][j],dp[t][i]+dis[i][j]);
				}
			}
			if (find)
				for (i=1;i<=n;i++)
					ans=min(ans,dp[t][i]);
		}
		if (ans==INF) ans=-1;
		cout<<ans<<endl;
	}
	return 0;
}