HDU 3001 Travelling(DP状态压缩)

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

 

Sample Output

100907 
题意：每个城市游览次数d>=1&&d<=2问最小花费。
三进制状态转移：dp[status+num[j]][j]=min(dp[status+num[j]][j],dp[status][i]+dis[i][j])；
#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<limits.h>typedef long long LL;using namespace std;#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )#define CLEAR( a , x ) memset ( a , x , sizeof a )const int INF=0x3f3f3f;int dp[60000][11];//记录i状态下以j为终点的值int f[60000][11];//记录i状态下j点的访问次数int num[12];int dis[15][15];int n,m;void init(){    num[1]=1;    REPF(i,2,11)        num[i]=num[i-1]*3;    for(int i=0;i<num[11];i++)    {        int temp=i;        REPF(j,1,10)        {            f[i][j]=temp%3;            temp/=3;        }    }//    for(int i=1;i<=10;i++)//       cout<<"2333  "<<f[12][i]<<endl;}int main(){    int x,y,val;    init();    while(~scanf("%d%d",&n,&m))    {        CLEAR(dp,INF);        CLEAR(dis,INF);//        cout<<"23333 "<<INF<<endl;        while(m--)        {            scanf("%d%d%d",&x,&y,&val);            if(dis[x][y]>val) dis[x][y]=dis[y][x]=val;        }        REPF(i,1,n)           dp[num[i]][i]=0;//        cout<<"2333  "<<dp[3][2]<<" "<<dp[2][3]<<endl;        int ans=INF;        for(int status=1;status<num[n+1];status++)        {            int ok=1;            REPF(i,1,n)            {                if(f[status][i]==0)//在状态status下的i次数为0，ok=0，还有地方没有游                    ok=0;                if(dp[status][i]==INF)//不存在这条道路                    continue;                REPF(j,1,n)                {                    if(i==j)  continue;                    if(f[status][j]==2)  continue;//状态status下j点走过两次                    if(dis[i][j]==INF)  continue;                    dp[status+num[j]][j]=min(dp[status+num[j]][j],dp[status][i]+dis[i][j]);//状态转移                }            }            if(ok)            {                REPF(i,1,n)                   ans=min(ans,dp[status][i]);            }        }        if(ans==INF)   ans=-1;        printf("%d\n",ans);    }    return 0;}