Football


Time Limit: 1000MS

 

Memory Limit: 65536K

Total Submissions: 4344

 

Accepted: 2220


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 = [pij] such that pij 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 pij. The matrix P will satisfy the constraints that pij = 1.0 − pji for all i ≠ j, and pii = 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

题意：

给你2^n个球队，必须相邻的球队才可以比赛！在告诉你i球队战胜j球队的概率，问最后哪支球队赢得概率最大！

思路:

概率DP，借鉴了别人的博客！（在判断球队是否相邻用到的位元算很巧妙~）

因为2^n个球队 需要n大轮比赛才能决定冠军！

因此，可以用dp[i][j]，表示第i大轮比赛，j球队赢得概率！

先遍历比赛轮数i，在遍历j，在遍历k，k表示j可以战胜的球队！

当判断j和k相邻时（可以打比赛），

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

表示在上一轮中，j和k都存活了下来，并且在这一轮中j战胜了k。

这样就解决了！

那么 如何判断两个球队是否相邻呢！

用到了^运算符，有一个性质 (2n) ^ (1) = 2n+1;  (2n+1) ^ (1) = 2n

因此先给每一个数 >> (i-1)，在进行^运算！就可以判断是否相邻了。

这个说不太好说明，写一下就很明了了！


一定要注意在进行^运算时 优先级问题，因为这个WA了两遍！

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn = 128 + 5;
double p[maxn][maxn];
double dp[maxn][maxn];
int main(){
	int n;
	while(scanf("%d",&n) == 1 && n != -1){
		int len = 1 << n;//球队的总数量！2^n； 
		for (int i = 1; i <= len; ++i)
			for (int j = 1; j <= len; ++j)
				scanf("%lf",&p[i][j]);
		memset(dp,0,sizeof(dp));
		for (int i = 1; i <= len; ++i)dp[0][i] = 1.0;//刚开始比赛时，所有队伍都未淘汰！
		for (int i = 1; i <= n; ++i){//正在进行第i次（大场）比赛！
			for (int j = 1; j <= len; ++j){
				for (int k = 1; k <= len; ++k){
					if ((((j-1) >> (i-1)) ^ 1) == ((k-1) >> (i-1)))//判断能否进行比赛（是否相邻！）  注意优先级问题~~ 
						dp[i][j] += dp[i-1][j] * dp[i-1][k] * p[j][k];
				}
			} 	
		}
		double MAX = 0.0;
		int ans_x;
		for (int i = 1; i <= len; ++i){
			if (dp[n][i] > MAX){
				MAX = dp[n][i];
				ans_x = i;
			}
		}
		printf("%d\n",ans_x);	
		//printf("%lf\n",MAX);
	}
	return 0;
}