POJ 3071 Football(概率dp)

Football

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 4816 Accepted: 2443

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, the jth 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

20.0 0.1 0.2 0.30.9 0.0 0.4 0.50.8 0.6 0.0 0.60.7 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.

题意：有n^2只足球队要进行n场足球比赛，按队伍编号进行比赛（编号为1，2，3，4，5，6，7，8；第一轮比赛为1vs2,3vs4,5vs6，7vs8）；每轮比赛的胜者在于对应的胜者进行比赛，知道决出总冠军。现在给出i队战胜j队的几率表（p[i][j]表示i队战胜j队的概率），问哪只队伍最有可能获得总冠军？

题解：我们用dp[i][j]表示在第i轮第j只队伍获胜的概率。 则可以得到状态转移方程为 dp[i][j]=dp[i-1][j]*sum；sum=sigma(dp[i-1][k]*p[j][k])。我们要枚举出在第i轮j所有的可能对手k，并计算出k在第i-1轮获胜的概率与j战胜k的乘积，全部求和得到sum。

代码如下：

nclude<cstdio>#include<cstring>#include<algorithm>using namespace std;#define maxn 200double p[maxn][maxn];double dp[maxn][maxn];//dp[i][j]表示在第i局第j队获胜的概率 int main(){int n,i,j,k;while(scanf("%d",&n)&&(n!=-1)){int m=(1<<n);for(i=0;i<m;++i){for(j=0;j<m;++j)scanf("%lf",&p[i][j]);}memset(dp,0,sizeof(dp));for(i=0;i<m;++i)dp[0][i]=1;for(i=1;i<=n;++i){for(j=0;j<m;++j){for(k=0;k<m;++k){if((j>>(i-1)^1)==(k>>(i-1)))//这里可以看成一颗二叉树，(j>>(i-1)^1)==(k>>(i-1))就表示(j>>(i-1))与(k>>(i-1))是同一个父节点的两个孩子节点，胜出者就能跟新到父节点 dp[i][j]+=dp[i-1][j]*dp[i-1][k]*p[j][k];}}}int ans=0;for(i=0;i<m;++i){if(dp[n][i]>dp[n][ans])ans=i;}printf("%d\n",ans+1);}return 0;}