POJ3071:Football(概率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

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.

 

题意：有2^n支队伍进行比赛，每行给出这支队伍打败各支队伍的几率，求出那支队伍获胜几率最大

思路：

概率DP，递推式：dp[i][j]=sigma(dp[i-1][j]*p[j][k]*dp[i-1][k])，然后判断两支队伍是否相邻

dp[i][j]代表第i轮中，第j支队伍存活的几率

#include <stdio.h>

#include <string.h>

#include <algorithm>

using namespace std;



double a[150][150],dp[150][150];



int main()

{

    int n,m,i,j,k,ans;

    while(~scanf("%d",&n),n+1)

    {

        m = 1<<n;

        for(i = 0; i<m; i++)

            for(j = 0; j<m; j++)

                scanf("%lf",&a[i][j]);

        memset(dp,0,sizeof(dp));

        for(i = 0; i<m; i++)//一开始所有队伍都存活

            dp[0][i] = 1;

        for(i = 1; i<=n; i++)

            for(j = 0; j<m; j++)

                for(k = 0; k<m; k++)

                {

                    int p = k>>(i-1),q = j>>(i-1);//判断是否相邻

                    if(p%2)

                    {

                        p--;

                        if(p==q)//判断p前进或后退以为，p==q的话证明他们是相邻的，进行比赛

                            dp[i][j] += dp[i-1][j]*dp[i-1][k]*a[j][k];//j存活，k存活，j打败k

                    }

                    else

                    {

                        p++;

                        if(p==q)

                            dp[i][j] += dp[i-1][j]*dp[i-1][k]*a[j][k];

                    }

                }

        ans = 0;

        for(i = 0; i<m; i++)//找最大

        {

            if(dp[n][ans]<dp[n][i])

                ans = i;

        }

        printf("%d\n",ans+1);

    }



    return 0;

}