PAT 甲级 1056 Mice and Rice_1056 mice and rice-优快云博客

本文链接：https://blog.youkuaiyun.com/a845717607/article/details/86707364

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流程和排名规则，并提供了一段AC代码示例，使用队列来模拟淘汰赛制，最终确定每位程序员的排名。

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1056 Mice and Rice （25 point(s)）

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

经验总结：

这一题，就.....首先你得弄懂题目到底是怎么进行比赛以及如何进行排名的，如果这个弄懂了，这一题就不难了，不过难就难在弄懂上面，首先，排名是这样的，在同一轮被决定了排名的，拥有相同的排名，具体排名根据前面的人数决定（比如前面有n个人，那么这一批人的排名就为n+1），至于实现方法，用队列比较适合模拟这种淘汰类型的比赛，具体实现细节见代码~

AC代码

#include <cstdio>
#include <cmath>
#include <cstring>
#include <queue>
using namespace std;
const int maxn=1010;
struct mouse
{
	int rank,weight;
}mice[maxn];
int main()
{
	int Np,Ng,t;
	scanf("%d %d",&Np,&Ng);
	for(int i=0;i<Np;++i)
		scanf("%d",&mice[i].weight);
	queue<int> q;
	for(int i=0;i<Np;++i)
	{
		scanf("%d",&t);
		q.push(t);
	}
	int temp=Np,group;
	while(q.size()!=1)
	{
		if(temp%Ng==0)
			group=temp/Ng;
		else
			group=temp/Ng+1;
		for(int i=0;i<group;++i)
		{
			int x=q.front();
			for(int j=0;j<Ng;++j)
			{
				if(i*Ng+j>=temp)
					break;
				int front=q.front();
				if(mice[front].weight>mice[x].weight)
				{
					x=front;
				}
				mice[front].rank=group+1;
				q.pop();
			}
			q.push(x);
		}
		temp=group;
	}
	mice[q.front()].rank=1;
	for(int i=0;i<Np;++i)
		printf("%d%c",mice[i].rank,i<Np-1?' ':'\n');
	return 0;
}