poj3229 哈密顿回路

很简单的题目，先将路径floyed处理，然后状态压缩DP即可。

ACcode：

#include<cstdio>
#include<cstring>

const int NS=15;
const int MS=1<<NS;
const double INF=1e12;

int n,m,x,y,f;
int ms,lim,res;
int dt[2]={80,120};
double tmp,k,z;
double time[NS];
double g[NS][NS];
double dp[MS][NS];

double Min(double a1,double b1)
{
    return a1<b1?a1:b1;
}

double Max(double a1,double b1)
{
    return a1>b1?a1:b1;
}

int getone(int st)
{
    int cnt=0;
    for (int i=st;i;i=i&(i-1)) cnt++;
    return cnt;
}

int main()
{
    while (~scanf("%d%d%lf",&n,&m,&k)&&(n+m+k))
    {
        ms=0,k*=12,res=-1,lim=1<<n;
        for (int i=0;i<m;i++)
        scanf("%d",&x),x--,ms|=1<<x;
        for (int i=0;i<n;i++)
        scanf("%lf",&time[i]);
        for (int i=0;i<n;i++)
        for (int j=0;j<n;j++)
        g[i][j]=i==j?0.0:INF;
        while (~scanf("%d%d%lf%d",&x,&y,&z,&f)&&(x+y+z+f))
        {
            x--,y--,z/=dt[f];
            g[y][x]=g[x][y]=Min(g[x][y],z);
        }

        for (int c=0;c<n;c++)
        for (int a=0;a<n;a++)
        for (int b=0;b<n;b++)
        g[a][b]=Min(g[a][b],g[a][c]+g[c][b]);

        for (int i=0;i<lim;i++)
        for (int j=0;j<n;j++)
        dp[i][j]=INF;
        for (int i=0;i<n;i++)
        dp[1<<i][i]=g[0][i]+time[i];

        for (int i=1;i<lim;i++)
        {
            for (int j=0;j<n;j++)
            {
                if (i&(1<<j))
                {
                    y=i^(1<<j);
                    for (int u=0;u<n;u++)
                    if (y&(1<<u))
                    dp[i][j]=Min(dp[i][j],dp[y][u]+g[u][j]+time[j]);
                }
//                printf("dp[%d][%d]=%.2lf\n",i,j,dp[i][j]);
            }
            if (((i&ms)==ms)&&(dp[i][0]<k*1.0)&&getone(i)>res)
            res=getone(i);
        }
        if (res<0) printf("No Solution\n");
        else printf("%d\n",res);
    }
    return 0;
}