POJ 3071-Football（概率dp）

最新推荐文章于 2019-10-02 23:48:46 发布

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本文介绍了一种使用概率动态规划的方法来预测特定足球淘汰赛中哪支球队最有可能赢得最终冠军。该方法通过构建一个概率矩阵并运用动态规划算法，考虑了各支球队之间的胜负概率。

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

概率dp。

题意 ： 足球淘汰赛。一共有n轮，共有2^n支队伍参赛，所以总共进行n轮比赛即可决出冠军。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。

然后下面的问题是如何求出k。可以列出在2进制状态下的比赛过程，然后可以发现规律：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\n",ans);
  }
  return 0;
}