Travelling
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3905 Accepted Submission(s): 1234
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
网上说是DP+状压，结果用bfs+状压水过了。。
开始时把每一个点都入队，模拟3进制处理每个状态，最后+优化。
#include
#include
#include
#include
#include
#include
#include
using namespace std; #define N 12 #define LL long long const int inf=0x3fffffff; int g[N][N]; int n,m,ans; int mark[N][60000]; struct node { int x,t,s,cnt; //位置、时间、状态、个数 friend bool operator<(node a,node b) { return a.t>b.t; } }; int gettmp(int x,int k) //得到X在3进制下的第K位是多少 { int t; while(x) { t=x%3; k--; if(k==0) break; x/=3; } return k?0:t; } void inti() //初始化数组 { int i,j; for(i=1;i<=n;i++) { for(j=0;j<(int)pow(3,n);j++) mark[i][j]=inf; } } void bfs() { int i; priority_queue
q; node cur,next; for(i=1;i<=n;i++) { cur.x=i; cur.s=pow(3,(i-1)); cur.t=0; cur.cnt=1; q.push(cur); mark[i][0]=0; } while(!q.empty()) { cur=q.top(); q.pop(); for(i=1;i<=n;i++) { if(g[cur.x][i]==inf) continue; next.cnt=cur.cnt; next.s=cur.s; next.t=cur.t+g[cur.x][i]; if(ans
=2) continue; next.s+=pow(3,(i-1)); if(t==0) { next.cnt++; if(next.cnt==n) { ans=min(ans,next.t); continue; } } if(next.t