POJ 3071-Football（概率dp）

POJ 3071-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, thejth 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

概率dp。

題意 ： 足球淘汰賽。一共有n輪，共有2^n支隊伍參賽，所以總共進行n輪比賽即可決出冠軍。設dp[i][j]為第i輪比賽中j隊勝出的概率。則dp[i][j]=dp[i-1][j]*dp[i-1][k]*p[j][k].k為在本輪j的對手。

然後下面的問題是如何求出k。可以列出在2進制狀態下的比賽過程，然後可以發現規律：j>>i-1^1==k>>i-1;

#include 
#include 
#include 
#include 
using namespace std;
#define ll long long
double dp[8][130],p[130][130];
int main()
{
  int n;
  while(scanf("%d",&n)!=EOF&&n!=-1)
  {
  	int num=1<>(i-1)^1)==(k>>i-1))
				dp[i][j]+=dp[i-1][j]*dp[i-1][k]*p[j][k];
	    }
	int ans;double Max=-1;
	for(int i=0;iMax)
		{
			ans=i+1;
			Max=dp[n][i];
		}
	}
	printf("%d\n",ans);
  }
  return 0;
}