zoukankan      html  css  js  c++  java
  • 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 = [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, thejth 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
    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。

  • 相关阅读:
    C# 调用AForge类库操作摄像头
    Composer简介及使用实例
    asp.net mvc 接入美圣短信 验证码发送
    敏捷模式下携程的接口自动化平台演变
    17个Python的牛逼骚操作,你都OK吗?
    如何打造一份优雅的简历?
    谈谈少儿编程
    打基础一定要吃透这12类 Python 内置函数
    求职日记丨秋招面试零失败,我拿下宝洁、华为、壳牌等offer
    我只想找个测试岗,你却百般刁难我!
  • 原文地址:https://www.cnblogs.com/mengfanrong/p/4688930.html
Copyright © 2011-2022 走看看