poj 3071 Football (概率DP水题)-优快云博客

本文探讨了一个涉及2^n个队伍的单淘汰赛制足球锦标赛，通过构建概率矩阵预测最有可能赢得整个锦标赛的队伍。采用动态规划算法计算每轮比赛中各队获胜的概率，并最终确定冠军热门。

G - Football

Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

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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 2ⁿvalues; 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.

题目：有2^n个队，相邻的两两打淘汰赛，，求最后哪个队夺冠的概率最大。

题解：dp[i][j]表示第i轮的时候，第j去支队伍赢的概率。那么dp[i][j]的前提就是i-1轮的时候，

　　　j是赢的，而且第i轮赢了对方接下来就是找到第i轮的时候，他的可能队手

　　　通过二进制可以发现规律，所有高位是一样的，第i位刚好相反，

　　　所以用位运算可以巧妙解决，见代码

　　　dp[i][j]=sigma(dp[i-1][j]*dp[i-1][k]*p[j][k])

　　　每经过一轮会淘汰掉一半的人，所以可以右移一位，即除以二求概率。

　　　具体的解释请看代码注释。O(∩_∩)O

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
const int maxn=200;
double dp[8][maxn];
double p[200][200];
int main()
{
    int n,ans;
    while(scanf("%d",&n)&&n!=-1)
    {
        memset(dp,0,sizeof(dp));
        int x=1<<n; //左移  2^n支队伍
        for(int i=0; i<x; i++)
            for(int j=0; j<x; j++)
                scanf("%lf",&p[i][j]);
        for(int j=0; j<x; j++)//初始化
            dp[0][j]=1;
        for(int i=1; i<=n; i++)//n次比赛
            for(int j=0; j<x; j++)//注意这里跑的核心是j 这个代码相对于j来说的
                for(int k=0; k<x; k++) // 复杂度最高7*2^14=114672 不会超时
                    //当然这里是可以优化的 这么写代码更短
                    if((j>>(i-1)^1)==(k>>(i-1)))//右移 多加括号 代表每次淘汰一半人
                        //这个if的意思是奇数只和上一个数比而偶数只和下一个数比
                        //异或一可以让奇数减一 偶数加一
                        dp[i][j]+=dp[i-1][j]*dp[i-1][k]*p[j][k];//注意k在前i-1轮也得赢
                        //别忘了加号 是把概率加起来
        double ans=-1;
        int ansj=0;
        /*for(int j=0;j<(1<<n);j++)
            cout<<dp[n][j]<<endl;*/ //输出各队胜率
        for(int j=0; j<x; j++) //选择胜率最高的那个队伍
            if(dp[n][j]>ans)
            {
                ans=dp[n][j];
                ansj=j+1;
            }
        printf("%d\n",ansj);
    }

    return 0;
}