传送门HDU3001

描述

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.

输入

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.

输出

Output the minimum fee that he should pay,or -1 if he can’t find such a route.

样例

Input2 11 2 1003 21 2 402 3 503 31 2 31 3 42 3 10

Output100907

题解

题意：从任意起点开始走完所有的点，每个点最多走两次，求最短路径与经典TSP问题不同的是，这里每个点可以走两次，所以采用三进制状压，012分别表示没走过，走过一次，走过两次。与二进制状压不同的是，我们不能通过位运算很方便地知道每一位的数值或者状态转移，所以我们要预处理出来每种状态的三进制下每一位的数值。dp[i][j]表示用状态i下以j为最后一个到达的点的最短距离，dp[sta+three[j]][j]=min(dp[sta+three[j]][j],dp[sta][i]+mp[i][j]);

Code

123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657#include<bits/stdc++.h>#define INIT(a,b) memset(a,b,sizeof(a))#define LL long longusing namespace std;const int inf=0x3f3f3f3f;const int N=20;const int mod=1e9+7;int n,m,vis[N];int three[11],dig[60000][20];//三进制数，每一个数的每一位（0,1,2）int mp[N][N],ans,dp[60000][20];//dp[i][j]第i种状态以j为结尾void init(){ three[0]=1; for(int i=1;i<=10;i++) three[i]=three[i-1]*3; int tem; for(int i=0;i<three[10];i++){ tem=i; for(int j=0;j<10;j++){ dig[i][j]=tem%3; tem/=3; } }}int main(){ int a,b,c; init(); while(~scanf("%d%d",&n,&m)){ INIT(mp,inf);INIT(dp,inf); while(m--){ scanf("%d%d%d",&a,&b,&c); mp[a-1][b-1]=mp[b-1][a-1]=min(mp[a-1][b-1],c); } //初始化第一个走的点 for(int i=0;i<n;i++) dp[three[i]][i]=0; int ans=inf,flag=1; for(int sta=0;sta<three[n];sta++){ flag=1; for(int i=0;i<n;i++){ if(dig[sta][i]==0)flag=0;//如果有0说明还有点没走过 if(dp[sta][i]==inf)continue;//如果当前状态i点没走过，则不可用 for(int j=0;j<n;j++){ if(mp[i][j]==inf||dig[sta][j]==2)continue;//通过当前状态推导下一个状态，如果当前状态下j已经走过两次了，那么下一个状态必然不能再到j dp[sta+three[j]][j]=min(dp[sta+three[j]][j],dp[sta][i]+mp[i][j]);//sta+three[j]相当于第j位+1 } } if(!flag)continue; //当存在所有点都走过的状态时就更新答案 for(int i=0;i<n;i++) ans=min(ans,dp[sta][i]); } if(ans==inf) printf("-1\n"); else printf("%d\n",ans); } return 0;}