poj3071Football(概率期望dp)

Football

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5620		Accepted: 2868

Description

Consider a single-elimination football tournament involving 2ⁿ teams, denoted 1, 2, …, 2ⁿ. In each round of the tournament, all teams still in the tournament are placed in a list in order of increasing index. Then, the first team in the list plays the second team, the third team plays the fourth team, etc. The winners of these matches advance to the next round, and the losers are eliminated. After n rounds, only one team remains undefeated; this team is declared the winner.

Given a matrix P = [p_ij] such that p_ij is the probability that team i will beat team j in a match determine which team is most likely to win the tournament.

Input

The input test file will contain multiple test cases. Each test case will begin with a single line containing n (1 ≤ n ≤ 7). The next 2ⁿ lines each contain 2ⁿ values; here, the jth value on the ith line represents p_ij. The matrix P will satisfy the constraints that p_ij = 1.0 − p_ji for all i ≠ j, and p_ii = 0.0 for all i. The end-of-file is denoted by a single line containing the number −1. Note that each of the matrix entries in this problem is given as a floating-point value. To avoid precision problems, make sure that you use either the double data type instead of float.

Output

The output file should contain a single line for each test case indicating the number of the team most likely to win. To prevent floating-point precision issues, it is guaranteed that the difference in win probability for the top two teams will be at least 0.01.

Sample Input

2
0.0 0.1 0.2 0.3
0.9 0.0 0.4 0.5
0.8 0.6 0.0 0.6
0.7 0.5 0.4 0.0
-1

Sample Output

2

Hint

In the test case above, teams 1 and 2 and teams 3 and 4 play against each other in the first round; the winners of each match then play to determine the winner of the tournament. The probability that team 2 wins the tournament in this case is:

P(2 wins)

= P(2 beats 1)P(3 beats 4)P(2 beats 3) + P(2 beats 1)P(4 beats 3)P(2 beats 4) = p21p34p23 + p21p43p24 = 0.9 · 0.6 · 0.4 + 0.9 · 0.4 · 0.5 = 0.396.

The next most likely team to win is team 3, with a 0.372 probability of winning the tournament.

Source

Stanford Local 2006

 

/*
因为2^n个球队 需要n大轮比赛才能决定冠军！
因此，可以用dp[i][j]，表示第i大轮比赛，j球队赢得概率！
先遍历比赛轮数i，在遍历j，在遍历k，k表示j可以战胜的球队！
当判断j和k相邻时（可以打比赛），
dp[i][j] +=dp[i-1][j] * dp[i-1][k] * p[j][k];
表示在上一轮中，j和k都存活了下来，并且在这一轮中j战胜了k。
这样就解决了！
那么 如何判断两个球队是否相邻呢！
用到了^运算符，有一个性质 (2n) ^ (1) = 2n+1;  (2n+1) ^ (1) = 2n
因此先给每一个数 >> (i-1)，在进行^运算！就可以判断是否相邻了。
这个说不太好说明，写一下就很明了了！
*/
#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
using namespace std;

double dp[8][200];//dp[i][j]表示在第i场比赛中j胜出的概率
double p[200][200];
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        if(n==-1)break;
        memset(dp,0,sizeof(dp));
        for(int i=0;i<(1<<n);i++)
          for(int j=0;j<(1<<n);j++)
            scanf("%lf",&p[i][j]);
            //cin>>p[i][j];
        for(int i=0;i<(1<<n);i++)dp[0][i]=1;
        for(int i=1;i<=n;i++)//2^n个人要进行n场比赛
        {
            for(int j=0;j<(1<<n);j++)
            {
                int t=j/(1<<(i-1));
                t^=1;
                dp[i][j]=0;
                for(int k=t*(1<<(i-1));k<t*(1<<(i-1))+(1<<(i-1));k++)
                  dp[i][j]+=dp[i-1][j]*dp[i-1][k]*p[j][k];
            }
        }
        int ans;
        double temp=0;
        for(int i=0;i<(1<<n);i++)
        {
            if(dp[n][i]>temp)
            {
                ans=i;
                temp=dp[n][i];
            }
        }
        printf("%d
",ans+1);
    }
    return 0;
}