POJ 3071 Football（概率DP）

Football

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 1934		Accepted: 958

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

 

题意：2^n个队进行足球赛，每个队打败另外一个队都有一个概率。
问最后胜利的概率最大的是哪只球队。

概率DP。

设dp[i][j]表示第i场比赛第j个球队胜利的概率。

画图看看就可以得出公式了。

第一轮，(1,2)(3,4)(5,6)``````

很容易算出来。

第二轮的时候。

比如算3胜出的概率。首先是3在第一轮要胜出，同时要打败(1,2)种的胜者，这是全概率公式了。

/*

POJ 3071

题意：2^n个队进行足球赛，每个队打败另外一个队都有一个概率。

问最后胜利的概率最大的是哪只球队。



概率公式，dp算一下就可以了。

*/

#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场比赛中j胜出的概率

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个人要进行n场比赛

        {

            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;

}