HDU3001 Travelling —— 状压DP（三进制）

题目链接：http://acm.split.hdu.edu.cn/showproblem.php?pid=3001

Travelling

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8286    Accepted Submission(s): 2703

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

 

 

 

 

题解：

1.对于每一个点， 有三种状态：没有被访问、被访问一次、被访问两次。所以需要用三进制进行压缩。

2.dp[status][last]表示：在状态为status下（即走了多少个点，以及每个点走了多少次），最后走的是last点的最小花费。

3.对于一个已有状态：dp[s][j]，枚举每一个点k，如果这个点k不是j，而k、j有边，且k被访问的次数不超过2次，那么下一步就可以访问k点。

4.类似的题目：HDU4856

 

 

代码如下：

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #include <vector>
 #include <queue>
 #include <stack>
 #include <map>
 #include <string>
 #include <set>
 #define ms(a,b) memset((a),(b),sizeof((a)))
 using namespace std;
 typedef long long LL;
 const double EPS = 1e-;
 const int INF = 2e9;
 const LL LNF = 2e18;
 const int MAXN = +;

 int n, m;
 int g[MAXN][MAXN], c[MAXN], dp[][];

 int main()
 {
     c[] = ;
     for(int i = ; i<=; i++)  //计算三进制
         c[i] = c[i-]*;

     while(scanf("%d%d", &n, &m)!=EOF)
     {
         memset(g, -, sizeof(g));
         for(int i = ; i<=m; i++)
         {
             int u, v, w;
             scanf("%d%d%d", &u, &v, &w);
             u--; v--;
             if(g[u][v]!=-) w = min(w, g[u][v]);
             g[u][v] = g[v][u] = w;
         }

         memset(dp, -, sizeof(dp));
         for(int i = ; i<n; i++)
             dp[c[i]][i] = ;

         int ans = INF;
         for(int s = ; s<c[n]; s++)
         for(int j = ; j<n; j++)
         {
             if(dp[s][j]==-) continue;  //此状态不存在， 跳过

             int cnt = ;
             for(int k = ; k<n; k++)
             {
                 if(j!=k && g[j][k]!=- && (s/c[k])%< )
                 {
                     int tmp = s + c[k];     //递推出下一个状态
                     if(dp[tmp][k]==- || dp[tmp][k]>dp[s][j]+g[j][k])   //更新
                         dp[tmp][k] = dp[s][j] + g[j][k];
                 }

                 if((s/c[k])%) cnt++;   //统计状态s下走了多少个点
             }

             if(cnt==n) ans = min(ans, dp[s][j]);    //如果状态s下走完所有点， 则更新答案
         }
         printf("%d\n", ans==INF?-:ans);
     }
 }