poj3071 Football

Football

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 2546 Accepted: 1289

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.

Source

Stanford Local 2006

大概属于二叉树的记忆化搜索吧……额，不过我更喜欢归结它为水题。。

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;struct Tree{    double win[1<<9];};Tree tree[1<<12];double fight[1<<9][1<<9];int n,cnt;void DFS(int t,int deep){    memset(tree[t].win,0,sizeof(tree[t].win));    if (deep==n)    {        tree[t].win[cnt++]=1;        return;    }    DFS(2*t+1,deep+1);    DFS(2*t+2,deep+1);    for (int i=0;i<(1<<n);i++)    {        for (int j=0;j<(1<<n);j++)        {            tree[t].win[i]+=tree[2*t+1].win[i]*tree[2*t+2].win[j]*fight[i][j];            tree[t].win[i]+=tree[2*t+2].win[i]*tree[2*t+1].win[j]*fight[i][j];        }    }}int main(){    int i,j;    while(1)    {        scanf("%d",&n);        if (n==-1) break;        for (i=0;i<(1<<n);i++)        {            for (j=0;j<(1<<n);j++)            {                scanf("%lf",&fight[i][j]);            }        }        cnt=0;        DFS(0,0);        double p=0;        int idx=-1;        for (i=0;i<(1<<n);i++)        {            if (tree[0].win[i]>p)            {                idx=i;                p=tree[0].win[i];            }        }        printf("%d\n",idx+1);    }    return 0;}