POJ 3071 Football （機率DP）

Football

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3285		Accepted: 1680

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
題意簡述：一場足球淘汰賽有2^n只隊伍參加，編號爲1,2.....，2^n，按隊伍編號進行比賽，每輪比賽的勝者再與對應比賽的勝者進行比賽，輸掉一場比賽該隊伍即被淘汰。如今給出每支隊伍與其餘全部隊伍進行比賽勝利的機率，求最可能成爲冠軍的隊伍編號。


思路：在一輪比賽中某支隊伍獲勝的機率等於該支隊伍再上一輪比賽中獲勝的機率與打敗本輪對手的機率的乘積。因此，能夠以dp[i][j]表示第j支隊伍在第i輪比賽中獲勝的機率，則dp[i][j]=dp[i-1][j]*sum，sum=∑(dp[i-1][k]*m[j][k])，枚舉出本輪全部j可能的對手k，計算出k在上一輪比賽中獲勝的機率與j打敗k的機率的乘積，所有求和獲得sum，從而算出dp[i][j]，max(dp[n][j])中的j便是題目所要求的解。

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 double p[130][130];
 5 double dp[130][8];
 6 int main()
 7 {
 8     int n,i,j,ans;
 9     while(scanf("%d",&n)!=EOF)
10     {
11         if(n==-1)
12             break;
13         int m=(1<<n);
14         for(i=0;i<m;i++)
15         {
16             for(j=0;j<m;j++)
17             {
18                 scanf("%lf",&p[i][j]);
19             }
20             dp[i][0]=1;
21         }
22         for(i=0;i<n;i++)
23         {
24             ans=0;
25             for(j=0;j<m;j++)
26             {
27                 double sum=0;
28                 for(int k=(1<<i);k<(1<<(i+1));k++)
29                 {
30                     sum+=dp[k^j][i]*p[j][k^j];
31                 }
32                 dp[j][i+1]=dp[j][i]*sum;
33                 if(dp[j][i+1]>dp[ans][i+1])
34                     ans=j;
35             }
36         }
37         printf("%d\n",ans+1);
38     }
39     return 0;
40 }