PAT A 甲级 1056 Mice and Rice (25分)

最新推荐文章于 2021-03-04 10:05:57 发布

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最新推荐文章于 2021-03-04 10:05:57 发布

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本文介绍了一个基于淘汰赛制的编程竞赛《Mice and Rice》的解决方案，通过模拟每轮比赛中最重小鼠的胜利过程，最终确定所有参赛小鼠的排名。使用了C++编程语言，借助vector数据结构实现了一边削减内容一边记录名次的高效算法。

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1056 Mice and Rice (25分)

题意

重点就是第三行。以示例为例，第三行的内容分析一下是“ID6号、0号、8号分别在第1、2、3位，他们三位在第一局中互相对决”。

思路

用了两个保存指针的vetor，方便一边削减vector内容（也就是模拟淘汰）一边记录名次。

然后就是模拟淘汰赛——每只鼠和当前同组最重鼠比较，留下胜者并给败者赋rank、erase败者，重复这个操作到只剩一只鼠。

代码

#include<iostream>
#include<vector>

using namespace std;

typedef struct {
	int id;
	int weight;
	int rank;
}mice;

int main()
{
	int np, ng, i, ii, temp, max, num;
	cin >> np >> ng;
	mice* m;
	vector<mice*> v1;
	vector<mice*> v2;
	for (i = 0; i < np; i++)
	{
		m = new mice;
		m->id = i;
		cin >> m->weight;
		v1.push_back(m);
	}
	for (i = 0; i < np; i++)
	{
		cin >> temp;
		v2.push_back(v1[temp]);
	}
	while (v2.size() > 1)
	{
		temp = v2.size();
		for (i = 0; i < v2.size(); i++)
		{
			max = v2[i]->weight;
			for (ii = 0; ii < ng - 1 && i + 1 < v2.size(); ii++)
			{
				if (v2[i + 1]->weight > max)
				{
					max = v2[i + 1]->weight;
					v2[i]->rank = temp / ng + 2;
					if (temp%ng == 0)
						v2[i]->rank--;
					v2.erase(v2.begin() + i);
				}
				else
				{
					v2[i + 1]->rank = temp / ng + 2;
					if (temp%ng == 0)
						v2[i + 1]->rank--;
					v2.erase(v2.begin() + i + 1);
				}
			}
		}
	}
	v2[0]->rank = 1;
	cout << v1[0]->rank;
	for (i = 1; i < np; i++)
	{
		cout << " " << v1[i]->rank;
	}
	return 0;
}

题目

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
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