1056 Mice and Rice (25分)

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

 

题解：

一开始没看懂题意···感觉主要考察阅读理解ORZ。

比较迷惑的一点没有看出题干里有说保证每个老鼠重量不同啊？？？ 

所以感觉这题太啰嗦了，不喜欢

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

struct Mouse
{
	int w;
	int r;
};

int main() {
	int n, g, o;
	cin >> n >> g;
	Mouse m[1010];
	queue<int> q;
	for (int i = 0; i < n; i++)
	{
		cin >> m[i].w;
	}
	for (int i = 0; i < n; i++)
	{
		cin >> o;
		q.push(o);
	}
	int temp = n, group;
	while (q.size() != 1)
	{
		group = (temp % g == 0) ? (temp / g) : (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 (m[front].w > m[k].w) k = front;
				m[front].r = group + 1;
				q.pop();
			}
			q.push(k);
		}
		temp = group;
	}
	m[q.front()].r = 1;
	for (int i = 0; i < n; i++)
	{
		cout << m[i].r;
		if (i < n - 1) cout << ' ';
	}
    return 0;
}

 

来源：CSDN

作者：Niku_

链接：https://blog.csdn.net/ikunsaikou/article/details/104110347