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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

/*
因為2^n個球隊 需要n大輪比賽才能決定冠軍！
因此，可以用dp[i][j]，表示第i大輪比賽，j球隊贏得概率！
先遍歷比賽輪數i，在遍歷j，在遍歷k，k表示j可以戰勝的球隊！
當判斷j和k相鄰時（可以打比賽），
dp[i][j] +=dp[i-1][j] * dp[i-1][k] * p[j][k];
表示在上一輪中，j和k都存活了下來，並且在這一輪中j戰勝了k。
這樣就解決了！
那麽 如何判斷兩個球隊是否相鄰呢！
用到了^運算符，有一個性質 (2n) ^ (1) = 2n+1;  (2n+1) ^ (1) = 2n
因此先給每一個數 >> (i-1)，在進行^運算！就可以判斷是否相鄰了。
這個說不太好說明，寫一下就很明了了！
*/
#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場比賽中j勝出的概率
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個人要進行n場比賽
        {
            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\n",ans+1);
    }
    return 0;
}

poj3071Football(概率期望dp)