POJ3071---Football

Football
Time Limit: 1000MS
 
Memory Limit: 65536K
Total Submissions: 3267
 
Accepted: 1668
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 = [pij] such that pij 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 pij. The matrix P will satisfy the constraints that pij = 1.0 − pji for all i ≠ j, and pii = 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
확률 dp+법칙,
dp[i][j]로 하여금 i번째 개인이 j라운드를 이길 확률을 표시하다
그리고 j라운드에서 i와 만날 수 있는 모든 팀원들을 찾아라(이진법을 만들고 가장 높은 곳을 찾아라)

/*************************************************************************
    > File Name: POJ3071.cpp
    > Author: ALex
    > Mail: [email protected]
    > Created Time: 2014 12 20      13 20 58 
 ************************************************************************/

#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

const int N = 200;
double dp[N][10];
double mat[N][N];
int n;

int diff(int a, int b)
{
    for (int i = n; i >= 0; i--)
    {
        if ((a & (1 << i)) != (b & (1 << i)))
        {
            return i;
        }
    }
}

int main()
{
    while (~scanf("%d", &n))
    {
        if (n == -1)
        {
            break;
        }
        for (int i = 1; i <= (1 << n); i++)
        {
            for (int j = 1; j <= (1 << n); j++)
            {
                scanf("%lf", &mat[i][j]);
            }
        }
        for (int i = 1; i <= (1 << n); i++)
        {
            for (int j = 1; j <=n; j++)
            {
                dp[i][j] = 0.0;
            }
        }
        for (int i = 1; i <= (1 << n); i++)
        {
            dp[i][0] = 1.0;
        }
        for (int i = 1; i <= n; i++)
        {
            for (int j = 1; j <= (1 << n); j++)
            {
                for (int k = 1; k <= (1 << n); k++)
                {
                    if (j == k)
                    {
                        continue;
                    }
                    int p = diff(j - 1, k - 1);
                    if (p != i - 1)
                    {
                        continue;
                    }
                    dp[j][i] += dp[j][i - 1] * dp[k][i - 1] * mat[j][k];
                }
            }
        }
        int cnt;
        double ans = 0;
        for (int i = 1; i <= (1 << n); i++)
        {
            if (ans < dp[i][n])
            {
                ans = dp[i][n];
                cnt = i;
            }
        }
        printf("%d
", cnt);
    }
    return 0;
}