【PTA-A】1056 Mice and Rice (25 分)（队列）

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本文介绍了一个名为MiceandRice的编程竞赛，参赛者通过编写代码控制鼠标在地图上移动以吃尽可能多的大米。文章详细解释了比赛规则，包括随机决定的参赛顺序、每组比赛的设定以及最终排名的确定方式。输入包括程序员数量、每组最大参赛数、鼠标重量及初始比赛顺序，输出则是每位程序员的最终排名。

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 NP programmers. Then every NG 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 NG 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: NP and NG (≤1000), the number of programmers and the maximum number of mice in a group, respectively. If there are less than NG 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 NP distinct non-negative numbers Wi (i=0,⋯,NP−1) where each Wi is the weight of the i-th mouse respectively. The third line gives the initial playing order which is a permutation of 0,⋯,NP−1 (assume that the programmers are numbered from 0 to NP−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

#include<iostream>
#include<queue>
using namespace std;
struct Node {
	int weight;
	int rank;
}no[1010];
int main() {
	int n, g,order; cin >> n >> g;
	for (int i = 0; i < n; i++) {
		cin >> no[i].weight;
	}
	queue<int> q;
	for (int i = 0; i < n; i++) {
		cin >> order;
		q.push(order);
	}
	int temp = n, group;
	while (q.size() != 1) {
		if (temp % g == 0)group = temp / g;
		else group = temp / g + 1;
		for (int i = 0; i < group; i++) {
			int k = q.front();
			for (int j = 0; j < g; j++) {
				if (i * g+j >= temp)break;
				int front = q.front();
				if (no[front].weight > no[k].weight)
					k = front;
				no[front].rank = group + 1;
				q.pop();
			}
			q.push(k);
		}
		temp = group;
	}
	no[q.front()].rank = 1;
	for (int i = 0; i < n; i++) {
		cout << no[i].rank;
		if (i < n - 1)cout << " ";
	}
	return 0;
}