Football
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 2882   Accepted: 1466

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 nrounds, 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, the jth value on the ith line representspij. 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

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.

Time: 79MS
#include"iostream"
#include"cstdio"
#include"cstring"
using namespace std;
const int ms=;
double p[ms][ms];
double dp[ms][];
int main()
{
int n,i,j;
while(scanf("%d",&n),n!=-)
{
int m=(<<n);
for(i=;i<m;i++)
{
for(j=;j<m;j++)
scanf("%lf",&p[i][j]);
dp[i][]=;
}
int ans;
for(i=;i<n;i++)
{
ans=;
for(j=;j<m;j++)
{
double sum=;
for(int k=(<<i);k<(<<(i+));k++)
{
sum+=dp[k^j][i]*p[j][k^j];
}
dp[j][i+]=dp[j][i]*sum;
if(dp[j][i+]>dp[ans][i+])
ans=j;
}
}
printf("%d\n",ans+);
}
return ;
}

Football的更多相关文章

  1. POJ 3071 Football

    很久以前就见过的...最基本的概率DP...除法配合位运算可以很容易的判断下一场要和谁比.    from——Dinic算法                         Football Time ...

  2. Football Foundation (FOFO) TOJ 2556

    The football foundation (FOFO) has been researching on soccer; they created a set of sensors to desc ...

  3. 17111 Football team

    时间限制:1000MS  内存限制:65535K 提交次数:0 通过次数:0 题型: 编程题   语言: C++;C Description As every one known, a footbal ...

  4. CodeForces 432B Football Kit

     Football Kit Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u Subm ...

  5. Football(POJ3071)

    Football Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 3469   Accepted: 1782 Descript ...

  6. 16年大连网络赛 1006 Football Games

    题目链接:http://acm.hdu.edu.cn/contests/contest_showproblem.php?cid=725&pid=1006 Football Games Time ...

  7. 三分--Football Goal(面积最大)

    B - Football Goal Time Limit:500MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Su ...

  8. HDU5873:Football Games

    题目链接: Football Games 分析: 先将分数排序,然后 设当前队编号为p,设个指针为p+1,然后p>1,每次p-=2,指针右移一位p==1,指针指向的队-=1p==0,从指针开始到 ...

  9. Codeforces Gym 100425H H - Football Bets 构造

    H - Football BetsTime Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hust.edu.cn/vjudge/contest/ ...

随机推荐

  1. android判断当前网络状态及跳转到设置界面

    今天,想做这个跳转到网络设置界面, 刚开始用 intent = new Intent(Settings.ACTION_WIRELESS_SETTINGS); 不料老是出现settings.Wirele ...

  2. 如何阅读Java源码

    刚才在论坛不经意间,看到有关源码阅读的帖子.回想自己前几年,阅读源码那种兴奋和成就感(1),不禁又有一种激动.源码阅读,我觉得最核心有三点:技术基础+强烈的求知欲+耐心. 说到技术基础,我打个比方吧, ...

  3. 获取week of year的小程序

    #coding=utf8 import urllib, BeautifulSoup web=urllib.urlopen("http://whatweekisit.com/") s ...

  4. 让sublime text 2更好地支持Python

    SublimeCodeIntel: ~/.codeintel/config里加了python和pythonExtraPaths的路径(Mac): {"Python" : {&quo ...

  5. Codevs No.2144 砝码称重2

    2016-05-31 22:01:16 题目链接: 砝码称重2 (Codevs No.2144) 题目大意: 给定N个砝码,求称出M的重量所需砝码最小个数 解法: 贪心 使砝码数量最小,当然是每个砝码 ...

  6. openstack 网络

    物理节点hosts解析配置

  7. RHEL6.4 KVM 桥接上网的设置

    关闭网络管理器 chkconfig NetworkManager off  ##和桥接有冲突,要关闭 service NetworkManager stop   修改eth0为物理网口,br0为桥接网 ...

  8. Transform DataGrid 套用格式

    <table class="easyui-datagrid" title="Transform DataGrid" style="width:5 ...

  9. 【转】强大的vim配置文件,让编程更随意

    原文地址:http://www.cnblogs.com/ma6174/archive/2011/12/10/2283393.html 花了很长时间整理的,感觉用起来很方便,共享一下. 我的vim配置主 ...

  10. codeforce 630N Forecast

    N. Forecast time limit per test 0.5 seconds memory limit per test 64 megabytes input standard input ...