POJ 3071-Football（可能性dp）

Football

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3145		Accepted: 1591

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。

题意 ： 足球淘汰赛。一共同拥有n轮，共同拥有2^n支队伍參赛，所以总共进行n轮比赛就可以决出冠军。设dp[i][j]为第i轮比赛中j队胜出的概率。则dp[i][j]=dp[i-1][j]*dp[i-1][k]*p[j][k].k为在本轮j的对手。

然后以下的问题是怎样求出k。能够列出在2进制状态下的比赛过程。然后能够发现规律：j>>i-1^1==k>>i-1;

#include <iostream>
#include <cmath>
#include <cstdio>
#include <cstring>
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<<n;
  	memset(dp,0,sizeof(dp));
  	for(int i=0;i<num;i++)
	{
		for(int j=0;j<num;j++)
		scanf("%lf",&p[i][j]);
	    dp[0][i]=1;
	}
	for(int i=1;i<=n;i++)
		for(int j=0;j<num;j++)
		for(int k=0;k<num;k++)
	    {
		    if((j>>(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;i<num;i++)
	{
		if(dp[n][i]>Max)
		{
			ans=i+1;
			Max=dp[n][i];
		}
	}
	printf("%d
",ans);
  }
  return 0;
}

版权声明：本文博客原创文章。博客，未经同意，不得转载。