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| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 3285 | Accepted: 1680 |
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, the jth 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
题意简述：一场足球淘汰赛有2^n只队伍参加，编号为1,2.....，2^n，按队伍编号进行比赛，每轮比赛的胜者再与对应比赛的胜者进行比赛，输掉一场比赛该队伍即被淘汰。现在给出每支队伍与其他所有队伍进行比赛胜利的概率，求最可能成为冠军的队伍编号。
思路：在一轮比赛中某支队伍获胜的概率等于该支队伍再上一轮比赛中获胜的概率与战胜本轮对手的概率的乘积。所以，可以以dp[i][j]表示第j支队伍在第i轮比赛中获胜的概率，则dp[i][j]=dp[i-1][j]*sum，sum=∑(dp[i-1][k]*m[j][k])，枚举出本轮所有j可能的对手k，计算出k在上一轮比赛中获胜的概率与j战胜k的概率的乘积，全部求和得到sum，从而算出dp[i][j]，max(dp[n][j])中的j即是题目所要求的解。
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 double p[130][130]; 5 double dp[130][8]; 6 int main() 7 { 8 int n,i,j,ans; 9 while(scanf("%d",&n)!=EOF) 10 { 11 if(n==-1) 12 break; 13 int m=(1<<n); 14 for(i=0;i<m;i++) 15 { 16 for(j=0;j<m;j++) 17 { 18 scanf("%lf",&p[i][j]); 19 } 20 dp[i][0]=1; 21 } 22 for(i=0;i<n;i++) 23 { 24 ans=0; 25 for(j=0;j<m;j++) 26 { 27 double sum=0; 28 for(int k=(1<<i);k<(1<<(i+1));k++) 29 { 30 sum+=dp[k^j][i]*p[j][k^j]; 31 } 32 dp[j][i+1]=dp[j][i]*sum; 33 if(dp[j][i+1]>dp[ans][i+1]) 34 ans=j; 35 } 36 } 37 printf("%d\n",ans+1); 38 } 39 return 0; 40 }
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原文地址：http://www.cnblogs.com/caterpillarofharvard/p/4230985.html
