Massive Algorithms: POJ 3071 -- Football (Probability DP)

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

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.

http://www.cnblogs.com/zhj5chengfeng/archive/2013/03/01/2939444.html

题目大意

有 2^n 支足球队，编号 1~2^n，现在给出每支球队打败其他球队的概率，问哪只球队取得冠军的概率最大？

球队间的比赛规则是：在每轮比赛中，所有还在场的球队按照编号从小到大排序，然后第 2*k-1 支队和第 2*k 支队比赛，胜利者进入下一轮，直到最后只剩一支球队为止。

做法分析

简单的概率DP

观察到球队的数量是 2^n，这道题就可以用比赛的轮次来划分阶段了。最后总决赛的两支队伍一定是一支来自编号 1~2^(n-1)，另一只来自编号 2^(n-1)+1~2^n，也就是说，一支球队是前一半球队中的冠军，另一只球队是后一半球队中的冠军。

枚举每支队伍夺得总冠军的概率

https://blog.csdn.net/zwj1452267376/article/details/52121632
题解：我们用dp[i][j]表示在第i轮第j只队伍获胜的概率。则可以得到状态转移方程为 dp[i][j]=dp[i-1][j]*sum；sum=sigma(dp[i-1][k]*p[j][k])。我们要枚举出在第i轮j所有的可能对手k，并计算出k在第i-1轮获胜的概率与j战胜k的乘积，全部求和得到sum。

if((j>>(i-1)^1)==(k>>(i-1)))//这里可以看成一颗二叉树，(j>>(i-1)^1)==(k>>(i-1))就表示(j>>(i-1))与(k>>(i-1))是同一个父节点的两个孩子节点，胜出者就能跟新到父节点

http://blog.csdn.net/xuechelingxiao/article/details/38520105

DP方程为 dp[i][j] = ∑(dp[i-1][j]*dp[i-1][k]*p[j][k]); //dp[i][j]表示第 i 轮的时候，第 j 支队伍赢的概率。、

对于其中位运算，可以用一个二叉树来更直观的理解 (j>>(i-1))^1) 跟 (k>>(i-1)分别表示一个父节点的两个子节点，当(j>>(i-1))^1) == (k>>(i-1)时，表明两个子节点是竞争关系，胜利者将更新到复接点。

 4 int n;
 5 double p[1000][1000];
 6 double dp[10][1000];
 8 int main()
 9 {
10     while(~scanf("%d", &n) && n != -1){
11         for(int i = 0; i < (1<<n); ++i){
12             for(int j = 0; j < (1<<n); ++j){
13                 scanf("%lf", &p[i][j]);
14             }
15         }
16         memset(dp, 0, sizeof(dp));
18         for(int i = 0; i < (1<<n); ++i){
19             dp[0][i] = 1;
20         }
21         for(int i = 1; i <= n; ++i){
22             for(int j = 0; j < (1<<n); ++j){
23                 for(int k = 0; k < (1<<n); ++k){
24                     if(((j>>(i-1))^1) == (k>>(i-1)))
25                         dp[i][j] += dp[i-1][k]*dp[i-1][j]*p[j][k];
26                 }
27             }
28         }
30         double ans = -1;
31         int Ans = -1;
32         for(int i = 0; i < (1<<n); ++i){
33             if(dp[n][i] > ans){
34                 ans = dp[n][i];
35                 Ans = i;
36             }
37         }
38         printf("%d\n", Ans+1);
39     }
41     return 0;
42 }

定义状态：f[i][st][en] 表示第 i 支球队在第 st 到第 en 支队伍中夺冠的概率，不难写出如下的状态转移方程：

令：mid=(st+en)>>1;

如果 i 是位于[st, mid] 这个区间中的：

f[i][st][en] += f[i][st][mid]*f[j][mid+1][en]*p[i][j];

其中，j 为第 mid+1 到 en 支球队中的冠军，我们可以从 mid+1 一直枚举到 en，并求和

如果 i 是位于 [mid+1, en] 这个区间中的，类似的，我们有：

f[i][st][en] += f[i][mid+1][en]*f[j][st][mid]*p[i][j];

其中，j 是第 st 到 mid 支球队中的冠军，我们可以冲 st 一直枚举到 mid，并求和

最后的目标就是 f[i][1][2^n]

Read full article from 3071 -- Football

POJ 3071 -- Football (Probability DP)

题目大意

做法分析

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