PAT甲级-1056-最大优先队列

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​​ (≤1000), 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​​ (i=0,⋯,N​P​​−1) 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,⋯,N​P​​−1 (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

#include <stdio.h>
#include <algorithm>
#include <stdlib.h>
#include <malloc.h>
#include <string.h>
#include <iostream>
#include <queue>
#include <string>
#include <vector>
#include <map>
#include <stack>
using namespace std;
const int maxn = 10008;
const int INF = 0x3f3f3f3f;

int a[maxn];
int id[maxn];
int ank[maxn];
struct node {
	int id, vue;
	friend bool operator < (node a, node b) { return a.vue < b.vue; }
}ant[maxn];

int main() {
	std::ios::sync_with_stdio(false);
	int n, m;
	priority_queue<node> q;
	cin >> n >> m;
	for (int i = 0; i < n; i++) cin >> ant[i].vue, ant[i].id = i;
	for (int i = 0; i < n; i++) cin >> a[i];
	int group = (n%m == 0) ? (n / m) : (n / m + 1);

	//初赛复赛不放一起，覆盖ant的话会有问题
	for (int i = 0; i < group; i++) { //初赛
		for (int j = i*m; j < (i + 1)*m && j < n; j++) q.push(ant[a[j]]);
		id[i] = q.top().id; //最大优先度的数
		q.pop();
		while (!q.empty()) {
			ank[q.top().id] = group + 1;
			q.pop();
		}
		if (group == 1) ank[id[i]] = 1;
	}
	int len = 0;
	while (group != 1) { //循环复赛
		len = group;
		group = (len%m == 0) ? (len / m) : (len / m + 1);
		for (int i = 0; i < group; i++) {
			for (int j = i*m; j < (i + 1)*m && j < len; j++) q.push(ant[id[j]]);
			id[i] = q.top().id; //将一组中出线的id覆盖至原有id数组中
			q.pop();
			while (!q.empty()) {
				ank[q.top().id] = group + 1;
				q.pop();
			}
			if (group == 1) ank[id[i]] = 1;
		}
	}
	for (int i = 0; i < n; i++) {
		if (i == (n - 1)) cout << ank[i] << endl;
		else cout << ank[i] << " ";
	}
}