題目

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 $N_{}$

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

解題思路

　　題目大意： 給你 $N_P$個老鼠以及他們的重量 $W$，每 $N_G$個老鼠分一個組，不夠 $N_G$個數的單獨算一個組，比他們每個組的最大值，最大值進入下一輪的比較，同組其餘老鼠皆為淘汰，並與其他組同時被淘汰的老鼠排名一致，求所有老鼠的排名。
　　解題思路： 這道題的難點在於對資料進行分組，比較並進行排名，其中，老鼠的排名等於該輪比賽分組個數+1——本輪比較分group個組，意味著有group個優勝者，也就是說，其餘所有被淘汰的都在這group之後，即group+1.所以我們直接每輪的分組數即可。大題過程，如圖所示——
　　

#include <iostream>
#include <vector>
using namespace std;

int main(){
	int n, m, x;
	cin >> n >> m;
	vector<int> mice(n);
	vector<int> order(n);
	int rank[n];
	fill(rank,rank+n,0);
	
	for (int i = 0; i < n; ++i) {
		cin>>mice[i];
	}
	for (int i = 0; i < n; ++i) {
		cin>>order[i];
	}
	vector<int> res;
	int Max, idex, group;
	while (order.size()>1) {
		group = order.size() / m + 1;
		if ((order.size() % m != 0) ) ++group;
		for (int i = 0; i < order.size(); ++i) {
			if (i%m == 0) {
				Max = -1;
				if (i != 0) {
					res.push_back(idex);
				}
			}
			if (Max < mice[order[i]]) {
				Max = mice[order[i]];
				idex = order[i];
			}
			rank[order[i]] = group;
		}
		res.push_back(idex);
		order = res;
		res.clear();
	}
	rank[order[0]] = 1;
	for (int i = 0; i < n; ++i) {
		if (i != 0) cout << ' ';
		cout << rank[i];
	}
	return 0;
}

總結

　　這道題看起來不難，但是做起來挺難的，我做了好久……如果是在考場上，我可能拿不到分……甚是慚愧。