scoi 2008 && bzoj 1076 奖励关

来源：互联网发布：淘宝上一元秒杀骗局编辑：程序博客网时间：2024/06/06 00:16

传送门:http://www.lydsy.com/JudgeOnline/problem.php?id=3223
思路:15?好,状压,OＫ．
这是转移方程
if((s[k]&j)==s[k])
f[i][j]+=max(f[i+1][j],f[i+1][j|1<<(k-1)]+a[k]);
else f[i][j]+=f[i+1][j];
代码

#include <cstdio>#include <cmath>#include <iostream>#include <cstring>#include <algorithm>#define ll long long#pragma comment(linker, "/STACK:1024000000,1024000000")using namespace std;int k,n,a[105],s[105];double f[70000][20];int main(){    scanf("%d%d",&k,&n);    for (int i=1;i<=n;i++)    {        scanf("%d",&a[i]);        int x=0;        while (~scanf("%d",&x))        {            if (x==0) break;            s[i]|=1<<(x-1);        }    }    for (int i=k;i>=1;i--)    {        for (int j=0;j<1<<(n);j++)        {           f[i][j]=0.0;            for(int k=1;k<=n;k++)              if((s[k]&j)==s[k])                  f[i][j]+=max(f[i+1][j],f[i+1][j|1<<(k-1)]+a[k]);              else   f[i][j]+=f[i+1][j];              f[i][j]/=(double)n;          }    }    printf("%lf",f[1][0]);}

0 0