POJ 3071 Football （概率DP）

绝地灬酷狼 2022-08-21 11:59 136阅读 0赞

POJ 3071 Football （概率DP）：[http://poj.org/problem?id=3071][http_poj.org_problem_id_3071]

题面：

<table> 
 <tbody> 
 <tr> 
 <td>Time Limit:&nbsp;1000MS</td> 
 <td>&nbsp;</td> 
 <td>Memory Limit:&nbsp;65536K</td> 
 </tr> 
 <tr> 
 <td>Total Submissions:&nbsp;4389</td> 
 <td>&nbsp;</td> 
 <td>Accepted:&nbsp;2239</td> 
 </tr> 
 </tbody> 
</table>

Description

Consider a single-elimination football tournament involving 2*n* teams, denoted 1, 2, …, 2*n*. 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 2*n* lines each contain 2*n* values; here, the *j*th value on the *i*th 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

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:

<table> 
 <tbody> 
 <tr> 
 <td>P(2 wins)&nbsp;</td> 
 <td>=&nbsp;P(2 beats 1)P(3 beats 4)P(2 beats 3) +&nbsp;P(2 beats 1)P(4 beats 3)P(2 beats 4) =&nbsp;p21p34p23&nbsp;+&nbsp;p21p43p24 = 0.9 · 0.6 · 0.4 + 0.9 · 0.4 · 0.5 = 0.396.</td> 
 </tr> 
 </tbody> 
</table>

The next most likely team to win is team 3, with a 0.372 probability of winning the tournament.

题目大意：

2^n个球队进行比赛，给出两两球队之间比赛获胜的概率，使用淘汰制，求最后获胜的概率最大的队伍。

题目分析：

用DP\[i\]\[j\]表示在第i场比赛中j队胜出的概率，那么dp\[i\]\[j\]的前提就是i-1轮的时候，j是赢的，而且在第i轮赢了对方，则转为去第i轮中去找，通过二进制可以发现规律，所有的高位都是一样的，只有第i位恰好相反，所以用为运算即可解决，且DP公式如下：

dp\[i\]\[j\]+=dp\[i-1\]\[j\]\*dp\[i-1\]\[k\]\*p\[j\]\[k\].

代码实现：

#include <iostream>
 #include <cstdio>
 #include <cstring>
 
 using namespace std;
 
 double dp[8][200];///表示在第i场比赛中j胜出的概率
 double p[200][200];
 int main()
 {
 int n;
 while(scanf("%d",&n)!=EOF)
 {
 if(n==-1) break;
 memset(dp,0,sizeof(dp));
 for(int i=0;i<(1<<n);i++)
 {
 for(int j=0;j<(1<<n);j++)
 {
 scanf("%lf",&p[i][j]);
 }
 }
 for(int i=0;i<(1<<n);i++)///开始先都初始化为1
 {
 dp[0][i]=1;
 }
 for(int i=1;i<=n;i++)///2^n个人要进行比赛
 {
 for(int j=0;j<(1<<n);j++)
 {
 int t=j/(1<<(i-1));
 t^=1;
 dp[i][j]=0;
 for(int k=t*(1<<(i-1));k<t*(1<<(i-1))+(1<<(i-1));k++)
 {
 dp[i][j]+=dp[i-1][j]*dp[i-1][k]*p[j][k];
 }
 }
 }
 int ans;
 double temp=0;
 for(int i=0;i<(1<<n);i++)
 {
 if(dp[n][i]>temp)
 {
 ans=i;
 temp=dp[n][i];
 }
 }
 printf("%d\n",ans+1);
 }
 return 0;
 }

[http_poj.org_problem_id_3071]: http://http://poj.org/problem?id=3071