hdu 3071 ( 概率dp )

最新推荐文章于 2020-02-12 10:18:41 发布

柠檬与茉莉

最新推荐文章于 2020-02-12 10:18:41 发布

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分类专栏：概率dp 文章标签： acm 概率dp

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本文探讨了如何使用概率算法来预测在单场淘汰赛中哪支队伍最有可能获胜，涉及比赛规则、输入格式、算法实现及输出解析。

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题目链接：http://poj.org/problem?id=3071

Football

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3146		Accepted: 1592

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

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

思路：这一阵狂刷概率dp~

由淘汰赛的规则可以推出来要是有 2^n个队伍，那么就肯定有n场比赛；

那么 dp[i][j] 表示第i场比赛 j队能赢的概率；

#include <iostream>
#include <string.h>
#include <string>
#include <cstdio>
#include <cmath>
#include <algorithm>
const int N=(1<<7)+5;
using namespace std;
double dp[10][N];
double p[N][N];

int main()
{
    int n;
    while(cin>>n)
    {
      if(n==-1)break;
      int top=1<<n;
      for(int i=0;i<top;i++)
         for(int j=0;j<top;j++)
           scanf("%lf",&p[i][j]);
      memset(dp,0,sizeof(dp));
      for(int i=0;i<top;i++)dp[0][i]=1;

      for(int i=1;i<=n;i++)
         for(int j=0;j<top;j++)
            for(int k=0;k<top;k++)
            {
              if(((j>>(i-1))^1)!=k>>(i-1))continue;
              dp[i][j]+=dp[i-1][k]*dp[i-1][j]*p[j][k];
            }
     int flag=-1;
     double ans=-1;

     for(int i=0;i<top;i++)
     {
       if(dp[n][i]>ans)
       {
         ans=dp[n][i];
         flag=i;
       }
     }
     printf("%d\n",flag+1);
    }
    return 0;
}