POJ 3071 Football(概率DP)

题目链接

不1Y都对不住看过那么多年的球。dp[i][j]表示i队进入第j轮的概率,此题用0-1<<n表示非常方便。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <cmath>
#include <algorithm>
using namespace std;
double dp[501][501];
double p[501][501];
int main()
{
    int n,i,j,mod,c,k,ans;
    double maxz;
    while(scanf("%d",&n)!=EOF)
    {
        if(n < 0) break;
        for(i = 0;i < (1<<n);i ++)
        {
            for(j = 0;j < (1<<n);j ++)
            {
                scanf("%lf",&p[i][j]);
            }
        }
        for(i = 0;i < (1<<n);i ++)
        {
            dp[0][i] = 1;
        }
        for(i = 1;i <= n;i ++)
        {
            for(j = 0;j < (1<<n);j ++)
            dp[i][j] = 0;
        }
        for(i = 1;i <= n;i ++)
        {
            for(j = 0;j < (1<<n);j ++)
            {
                mod = j%(1<<i);
                if(mod >= (1<<(i-1)))
                {
                    c = (j/(1<<i))*(1<<i);
                    for(k = 0;k < (1<<(i-1));k ++)
                    {
                        dp[i][j] += dp[i-1][j]*dp[i-1][c+k]*p[j][c+k];
                    }
                }
                else
                {
                    c = (j/(1<<i))*(1<<i)+(1<<(i-1));
                    for(k = 0;k < (1<<(i-1));k ++)
                    {
                        dp[i][j] += dp[i-1][j]*dp[i-1][c+k]*p[j][c+k];
                    }
                }
            }
        }
        ans = 0;maxz = 0;
        for(i = 0;i < (1<<n);i ++)
        {
            if(maxz < dp[n][i])
            {
                ans = i;
                maxz = dp[n][i];
            }
        }
        printf("%d
",ans+1);
    }
    return 0;
}