poj3071 Football

学习位运算在比赛的技巧

http://poj.org/problem?id=3071

Football

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3075		Accepted: 1558

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, andp_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：

第一个队跟第二个对比赛，依次类推，赢了的比下一场。如图：

                            *

              *                    *

      *            *         *          *

  0     1     2     3     4      5       6     7

 00   01  10    11  100  101  110  111

可以看出

                   //每队只和其在当轮相邻的队伍进行比赛
                        //第一轮00和01可以比赛，01和10就不可以比赛（因为分组不同，见题意）
                        //第二轮00x和01x进行比赛
                   //第三轮0xx和1xx比赛，
推出  if(((j>>(i-1))^1) == (k>>(i-1)))
 //j和k可以进行比赛。                      

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

//dp[i][j]表示第i轮，第j队打赢的概率。

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
int n;
double a[200][200],dp[200][200];
int pow(int x)
{
    int i,sum=1;
    for(i=1;i<=x;i++)
        sum*=2;
    return sum;
}
int main()
{
    int i,j,m,k;
  while(~scanf("%d",&n))
   {
       memset(dp,0,sizeof(dp));
     if(n==-1)
        break;
        m=pow(n);
      for(i=0;i<m;i++)
        for(j=0;j<m;j++)
         scanf("%lf",&a[i][j]);
         for(i=0;i<m-1;i=i+2)
           {
              dp[1][i]=a[i][i+1];
            dp[1][i+1]=a[i+1][i];
           }
         for(i=2;i<=n;i++)
         {
             for(j=0;j<m;j++)
             {
                 for(k=0;k<m;k++)
                 {
                     if(((j>>(i-1))^1) == (k>>(i-1)))
                         dp[i][j] += dp[i-1][k]*dp[i-1][j]*a[j][k];
                 }
             }
         }
      double mmax=0;
      int team;
      for(i=0;i<m;i++)
       {
          // printf("dp=%lf\n",dp[n][i]);
         if(mmax<dp[n][i])
             {
                 mmax=dp[n][i];
                 team=i;
             }
       }
     printf("%d\n",team+1);
   }
   return 0;
}

 

转载于:https://www.cnblogs.com/cancangood/p/3929852.html