HDU3001——Travelling（状态压缩DP）

Travelling

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 2393    Accepted Submission(s): 692

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
 

解析：

         TSP问题，类似于我写的上一篇POJ3311。。。

         只是这题要用三进制。。。0表示还没走，1表示走了一次，2表示走了2次

         具体解析见代码   

代码：

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
#define inf 0x1f1f1f1f

int n,m,map[12][12];
int pos[12]={0,1,3,9,27,81,243,729,2187,6561,19683,59049};
int wei[60000][12]; //wei[s][i]表示状态s的第i位
int dp[12][60000]; //dp[i][s]表示i号点，状态为s

void read()
{
    freopen("hdu3001.in","r",stdin);
    freopen("hdu3001.out","w",stdout);
}

void work()
{
    for(int i=0;i<59050;++i) //预处理wei数组
    {
        int t=i;
        for(int j=1;j<=10;++j)
        {
            wei[i][j]=t%3;
            t/=3;
            if(t==0)break;
        }
    }
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        memset(map,inf,sizeof(map));
        while(m--)
        {
            int a,b,c;
            scanf("%d%d%d",&a,&b,&c);
            map[a][b]=map[b][a]=min(map[a][b],c);//处理重边
        }
        memset(dp,inf,sizeof(dp));
		for(int i=1;i<=n;i++)dp[i][pos[i]]=0;//递推边界
		int ans=inf;
		for(int s=0;s<pos[n+1];s++)
		{
			int vis=1;   //标记所有点是否全部走过
			for(int i=1;i<=n;i++)
			{
				if(wei[s][i]==0)vis=0; //有点没走过
				if(dp[i][s]==inf)continue;
				for(int j=1;j<=n;j++)                               //枚举的点要满足：j不等于i
				{                                                   //j没有走两次以上
					if(i==j||wei[s][j]>=2||map[i][j]==inf)continue;//i与j之间有边相连
					int S=s+pos[j];  //找出新状态(即s+3^j)
					dp[j][S]=min(dp[j][S],dp[i][s]+map[i][j]);//更新dp
				}
			}
            if(vis)   //所有点都走过
				for(int i=1;i<=n;i++)
					ans=min(ans,dp[i][s]);
		}
		if(ans==inf)printf("-1\n");else printf("%d\n",ans);
    }
}

int main()
{
    read();
    work();
    return 0;
}