poj 3071Football（概率dp）

最新推荐文章于 2018-03-19 08:09:00 发布

原创最新推荐文章于 2018-03-19 08:09:00 发布 · 437 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#acm #dp #概率dp #PKU

poj 同时被 2 个专栏收录

7 篇文章

订阅专栏

dp-概率dp

3 篇文章

订阅专栏

本文介绍了一种使用概率动态规划的方法来预测单淘汰制足球锦标赛中哪支球队最有可能赢得比赛。通过构建概率矩阵并采用动态规划算法，可以计算出每支球队赢得锦标赛的概率。

Football

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3340		Accepted: 1713

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

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

Stanford Local 2006

概率dp

dp[i][j]为在第i次比赛中j队获胜的概率

dp[i][j]=dp[i-1][j]*p0[i][j] p0[i][j]为第i轮获胜的概率

p0[i][j]=∑p[j][k]*dp[i-1][k]

关于k的取值，我们观察发现如果按照2^(i-1)划分区域，那么第1个区域的队伍在第i轮只可能和第2个区域的对战。

每轮对每一队所属区域标号为(i-1)/2^(i-1)+1

#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<stack>
#include<string>
#define MAX 1005
#define INF 1000000
using namespace std;
double dp[10][150],p[150][150],p0[10][150];
int main(void)
{
	int n,tm;
	double mx;
	int mxi;
	int i,j,k,l;
	int flag;
	while(scanf("%d",&n)&&(n+1))
	{
		memset(dp,0,sizeof(dp));
		memset(p,0,sizeof(p));
		memset(p0,0,sizeof(p0));
		tm=1<<n;
		for(i=1;i<=tm;i++)
			for(j=1;j<=tm;j++)
			scanf("%lf",&p[i][j]);
		for(i=1;i<=tm;i++)dp[0][i]=1;

        <span style="white-space:pre">	</span>for(i=1;i<=n;i++)
			for(j=1;j<=tm;j++)
		{
			flag=(j-1)/(1<<(i-1))+1;
			if(flag&1)l=flag*(1<<(i-1))+1;
			else l=(flag-2)*(1<<(i-1))+1;
			for(k=l;k<=l+(1<<(i-1))-1;k++)
				p0[i][j]+=p[j][k]*dp[i-1][k];
			dp[i][j]=dp[i-1][j]*p0[i][j];
		}
		mxi=1;mx=dp[n][1];
		for(i=2;i<=tm;i++)
			if(dp[n][i]>mx)
			{
				mx=dp[n][i];
				mxi=i;
			}
		printf("%d\n",mxi);
	}
	return 0;
}