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HDU 3001 Travelling

2024-07-14 21:25:01   228人阅读

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


题意：一个Acmer要出去旅行，有n个城市m条路，一个城市不能去2次以上，求经过所有城市一次的最短路。

思路：3进制计算，就是要开一个数组存3进制数了。

AC代码：

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <stdlib.h>

using namespace std;
const int INF=0x1f1f1f1f;
int mp[15][15];
int dp[59050][15];
int pow3[12]={0,1,3,9,27,81,243,729,2187,6561,19683,59049};
int n,m;
int p[59050][15];


int main(){
    for(int i=0;i<59050;i++){
        int temp=i;
        for(int j=1;j<=10;j++){
            p[i][j]=temp%3;
            temp/=3;
            if(temp==0) break;
        }
    }
    while(~scanf("%d%d",&n,&m)){
        memset(mp,INF,sizeof(mp));
        for(int i=0;i<m;i++){
            int a,b,c;
            scanf("%d%d%d",&a,&b,&c);
            if(c<mp[a][b])//竟然有多重边T_T,坑死我啦;
                mp[a][b]=mp[b][a]=c;
        }
        memset(dp,INF,sizeof(dp));
        for(int i=1;i<=n;i++) dp[pow3[i]][i]=0;
        bool flag;
        int ans=INF;
        for(int s=0;s<pow3[n+1];s++){
            flag=true;
            for(int i=1;i<=n;i++){
                if(p[s][i]==0) flag=false;
                if(dp[s][i]==INF) continue;
                for(int j=1;j<=n;j++){
                    if(i==j) continue;
                    if(p[s][j]>=2) continue;
                    if(mp[i][j]==INF) continue;
                    int ts=s+pow3[j];
                    dp[ts][j]=min(dp[ts][j],dp[s][i]+mp[i][j]);
                }
            }
            if(flag){
                for(int k=1;k<=n;k++){
                    ans=min(ans,dp[s][k]);
                }
            }
        }
        if(ans==INF) printf("-1\n");
        else printf("%d\n",ans);
    }
    return 0;
}






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