POJ 3071 Football（0）

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本文介绍了一种使用概率动态规划的方法来预测单淘汰制足球比赛中最有可能获胜的队伍。通过构建概率矩阵并运用DP算法，可以计算出各轮次中队伍继续获胜的概率，最终找出夺冠几率最高的队伍。

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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

解题思路：概率DP，dp[i][j]表示第i轮比赛结束后j仍获胜的概率,状态转移时我们需要确定在该轮比赛时哪些选手可能与j进行比赛，然后计算概率和即可。

第1轮结束后每两位中剩余一位，第2轮结束后每4位中剩余一位，因此很容易求得。

#include <ctime>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <string>
#include <vector>
#include <map>
#include <algorithm>
using namespace std;
const int maxn = 210;
double p[maxn][maxn];
double dp[10][maxn];

int main() {

    //freopen("aa.in", "r", stdin);

    int n;
    while(scanf("%d", &n) != EOF) {
        if(n == -1) break;
        for(int i = 1; i <= (1<<n); ++i) {
            for(int j = 1; j <= (1<<n); ++j) {
                scanf("%lf", &p[i][j]);
            }
        }
        for(int i = 1; i <= (1<<n); ++i) {
            dp[0][i] = 1.0;
        }
        for(int i = 1; i <= n; ++i) {
            for(int j = 1; j <= (1<<n); ++j) {
                int k = j / (1<<(i-1));
                if(j%(1<<(i-1))) k++;
                int s, e;
                if(k%2) {
                    s = k*(1<<(i-1)) + 1;
                    e = (k+1)*(1<<(i-1));
                } else {
                    s = (k-2)*(1<<(i-1)) + 1;
                    e = (k-1)*(1<<(i-1));
                }
                dp[i][j] = 0;
                for(int l = s; l <= e; ++l) {
                    dp[i][j] += dp[i-1][j] * dp[i-1][l] * p[j][l];
                }
            }
        }
        double ans = 0.0;
        int num = -1;
        for(int i = 1; i <= (1<<n); ++i) {
            if(ans < dp[n][i]) {
                ans = dp[n][i];
                num = i;
            }
        }
        printf("%d\n", num);
    }
    return 0;
}