POJ 3071

Football

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 720		Accepted: 330

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

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难度： 3   代码： 1      分类： 动态规划

分析：

其实hint中就已经提示dp的方程了，p[k][i]=p[k-1][i]*(sigma(p[k-1][j]*m[i][j])),p[k][i]为选手i在第k轮出现的概率，m[i][j]为概率矩阵中的元素，表示i胜j的概率；显然p[0][i]=1.0。问题集中到j的选取范围，易知第k轮涉及到的j的个数是2^(k-1)次，即1>>(k-1)，就只需要求出范围的开始就可以了。考察

0   0  1  2  3  4  5  6   7   

1   1  0  3  2  5  4  7   6

2   2  2  0  0  6  6  4   4

3   4  4  4  4  0  0  0   0

由这个可知上一次是序列的扩大2倍是下一轮的取值，即 1 0  3  2-> 2  2  0  0  6  6  4  4;

取得第一轮的取值是i^1,所以对于第k轮只需要扩展成 ((i>>(k-1))^1)<<(k-1)即可。

代码：

 

#include  < stdio.h >

double  m[ 128 ][ 128 ];
double  p[ 8 ][ 128 ];

int  main()
... {
    int n,size,i,j,k,start,num,best;
    
    while(scanf("%d",&n),n!=-1)
    ...{
        size=1<<n;
        for(i=0;i<size;i++)
            for(j=0;j<size;j++)
            ...{
                scanf("%lf",&(m[i][j]));
            }
        for(i=0;i<size;i++)
        ...{
            p[0][i]=1.0;
        }
        for(k=1;k<=n;k++)
        ...{
            for(i=0;i<size;i++)
            ...{
                p[k][i]=0.0;
                num=1<<(k-1);
                start=(((i>>(k-1))^1)<<(k-1));                
                for(j=start;j<num+start;j++)
                ...{
                    p[k][i]+=p[k-1][j]*m[i][j];
                }    
                p[k][i]*=p[k-1][i];
            }                
        }    
        best=0;
        for(i=1;i<size;i++)
        ...{
            if(p[n][i]>p[n][best])
            ...{
                best=i;                
            }
        }
        printf("%d ",best+1);        
    }
    return 0;
}