poj 3071 Football 概率动态规划

Football

Time Limit: 1000MS Memory Limit: 65536K

Total Submissions: 4841Accepted: 2459

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
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.

题目简述：

   有·2^n个队伍参加足球比赛，比赛采用淘汰赛的形式，所以一共要比n轮；第一次第0对跟第1队比，第2次第0队跟第2或3队比，以此类推；输入一个矩形表示各队之间获胜的概率，求获胜几率最大的那支队伍。

 思路：

    d[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]);枚举本轮j所以可能的对手k。

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>const  int maxn = 1<<8;using namespace std;float dp[8][maxn], a[maxn][maxn];int main() {    int i, j, k, n ,m;    while(scanf("%d", &n)&&(n+1)) {//这里wa了一次，不能写&&n，n=-1也会进循环。        m = 1<<n;        for(i = 0; i < m ;i++) {            for(j = 0; j < m; j++)                scanf("%f", &a[i][j]);            dp[0][i] = 1;        }            for(i = 0; i < n; i++)            for(j = 0; j < m; j++) {                float sum = 0;                for(k = 1<<i; k < (1<<(i+1)); k++) { //这两句话很厉害，这个循环保证了第i轮可以与j比赛的队伍个数                    sum += dp[i][j^k]*a[j][j^k];     //怎么讲呢，这里用的很巧妙，当i=0时，k可以等于1；当i=1时k可以等于2,3；                }                                    //当i=2时，k可以等于4,5,6,7；就拿i=2时来说，这时当j等于0到7时，                dp[i+1][j] = dp[i][j]*sum;        //只可能是0,1,2,3中的一支队伍与4,5,6,7之间的一支队伍比赛；            }                                      //当j分别等于0，1，2，3时；j^k都等于4,5,6,7；                                              //    当j分别等于4,5,6,7时；j^k有都等于0,1,2,3.没解释清，不想解释了(+﹏+)~            int p = 0;            for(i = 1; i < m; i++)                if (dp[n][i] > dp[n][p])                    p = i;            cout<<p+1<<endl;    }    return 0;}