PAT甲级——A1056 Mice and Rice

Mice and Rice is the name of a programming contest in which each programmer must write a piece of code to control the movements of a mouse in a given map. The goal of each mouse is to eat as much rice as possible in order to become a FatMouse.

First the playing order is randomly decided for N​P​​ programmers. Then every N​G​​ programmers are grouped in a match. The fattest mouse in a group wins and enters the next turn. All the losers in this turn are ranked the same. Every N​G​​ winners are then grouped in the next match until a final winner is determined.

For the sake of simplicity, assume that the weight of each mouse is fixed once the programmer submits his/her code. Given the weights of all the mice and the initial playing order, you are supposed to output the ranks for the programmers.

Input Specification:

Each input file contains one test case. For each case, the first line contains 2 positive integers: N​P​​ and N​G​​ (≤), the number of programmers and the maximum number of mice in a group, respectively. If there are less than N​G​​ mice at the end of the player's list, then all the mice left will be put into the last group. The second line contains N​P​​ distinct non-negative numbers W​i​​ (,) where each W​i​​ is the weight of the i-th mouse respectively. The third line gives the initial playing order which is a permutation of 0 (assume that the programmers are numbered from 0 to N​P​​−1). All the numbers in a line are separated by a space.

Output Specification:

For each test case, print the final ranks in a line. The i-th number is the rank of the i-th programmer, and all the numbers must be separated by a space, with no extra space at the end of the line.

Sample Input:

11 3
25 18 0 46 37 3 19 22 57 56 10
6 0 8 7 10 5 9 1 4 2 3

Sample Output:

5 5 5 2 5 5 5 3 1 3 5

 1 #include <iostream>
 2 #include <vector>
 3 #include <queue>
 4 using namespace std;
 5 int Np, Ng, w[1010], res[1010];
 6 int main()
 7 {
 8     cin >> Np >> Ng;
 9     for (int i = 0; i < Np; ++i)
10         cin >> w[i];
11     queue<int>q;
12     for (int i = 0; i < Np; ++i)
13     {
14         int a;
15         cin >> a;
16         q.push(a);
17     }
18     while (q.size() > 1)//即在产生冠军前需要继续迭代
19     {
20         int group = q.size() / Ng + (q.size() % Ng == 0 ? 0 : 1);
21         int size = q.size();
22         for (int i = 1; i <= group; ++i)
23         {
24             int lastN = i == group ? (size - (group - 1)*Ng) : Ng;//每组队员数量
25             int index = q.front();
26             for (int j = 0; j < lastN; ++j)
27             {
28                 res[q.front()] = group + 1;//loser的排名为分组数量 + 1
29                 index = w[index] > w[q.front()] ? index : q.front();
30                 q.pop();
31             }
32             q.push(index);//存下小组冠军的序号
33         }
34     }
35     res[q.front()] = 1;//产生冠军
36     for (int i = 0; i < Np; ++i)
37         cout << res[i] << (i == Np - 1 ? "" : " ");
38     return 0;
39 }