HDU3001(KB2-J 状态压缩dp)

Travelling

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8103    Accepted Submission(s): 2642

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

 

//2017-08-19
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int N = 15;
const int INF = 0x3f3f3f3f;
int G[N][N], n, m, pow3[N], dp[N][60000];//dp[i][S]表示到达节点i，状态为S时的最小费用

//询问节点i在状态S下经过了几次。将状态压缩为一个三进制数。
int query(int i, int S){
    return (S/pow3[i])%3;
}

int main()
{
    //预处理3的次方，pow3[n]表示3的n次方。
    pow3[0] = 1;
    for(int i = 1; i < N; i++)
          pow3[i] = pow3[i-1]*3;
    while(scanf("%d%d", &n, &m)!=EOF){
        int u, v, w;
        memset(G, INF, sizeof(G));
        memset(dp, INF, sizeof(dp));
        for(int i = 0; i < m; i++){
            scanf("%d%d%d", &u, &v, &w);
            u--; v--;//节点从0到n-1编号。
            G[u][v] = G[v][u] = min(G[u][v], w);//去重边
        }
        //初始化dp，因为每一个点都可以作为起点，所以到达i节点1次的最小费用为0。
        for(int i = 0; i < n; i++)
              dp[i][pow3[i]] = 0;
        int ans = INF;
        for(int S = 0; S < pow3[n]; S++){
            bool fg = true;
            for(int i = 0; i < n; i++){
                //检查是否每个节点都已经经过
                if(query(i, S) == 0){
                    fg = false;
                    continue;
                }
                //转移到下一个节点
                for(int v = 0; v < n; v++){
                    if(query(v, S) == 2)
                          continue;
                    dp[v][S+pow3[v]] = min(dp[v][S+pow3[v]], dp[i][S]+G[i][v]);
                }
            }
            if(fg){
                for(int i = 0; i < n; i++)
                      ans = min(ans, dp[i][S]);
            }
        }
        if(ans == INF)printf("-1
");
        else printf("%d
", ans);
    }

    return 0;
}