poj 3071 Football

题目：poj 3071 Football

tag：概率dp

思路：

Round1 :

                 0 play with 1

                 2 play with 3

                 4 play with 5

                 ...

Round 2:

               0 can play with 2 or 3    vice versa

               1 can play with 2 or 3    vice versa

               4 can play with 6 or 7    vice versa

               5 can play with 6 or 7    vice versa

               ...

Round i 

              player j can play with k   

             only   (j>>(i-1)) == ( (k>>(i-1))^1) 

#include <cstring>#include <algorithm>#include <cmath>#include <iostream>#include <cstdio>using namespace std;#define maxn ((1<<7)+1)double dp[8][maxn];double p[maxn][maxn];int main(){int n;while(scanf("%d",&n)!=EOF){if(n==-1)break;memset(dp,0,sizeof(dp));int m=(1<<n);for(int i=0;i<m;i++)dp[0][i]=1;for(int i=0;i<m;i++)for(int j=0;j<m;j++)scanf("%lf",&p[i][j]);for(int i=1;i<=n;i++)for(int j=0;j<m;j++){for(int k=0;k<m;k++){if((j>>(i-1)) == ((k>>(i-1))^1))dp[i][j]+=dp[i-1][k]*dp[i-1][j]*p[j][k];}}int ans=1;double mx=0;for(int i=0;i<m;i++){if(mx<dp[n][i]){mx=dp[n][i];ans=i;}}printf("%d\n",ans+1);}return 0;}