(PAT 1056) Mice and Rice (队列+模拟)

墨蓝 2022-03-25 05:17 345阅读 0赞

**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

解题思路:

这题想了一个下午,不过最后还是没有看题解做出来了

给出NP只老鼠的质量,根据第二列给出的序列排序,按这个顺序把这些老师按照NG分为一组,最后不够NG的单独分为一组

选出它们之中质量大的老鼠晋级,把所有晋级的老鼠再按照NG分组开始第二轮比较

直到最后晋级出唯一一只老鼠

把老鼠的序列入队,每次比较时按照 老鼠数组\[序列\].重量 这样的写法互相进行比较就行,不用再单独分一个数组出来

![watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2FsZXgxOTk3MjIy_size_16_color_FFFFFF_t_70][]

每轮分组的组数可以根据下面这个公式得到

int temp = NP;  
int group = temp % NG == 0 ? temp / NG : temp / NG + 1;

#include <iostream>
    #include <algorithm>
    #include <queue>
    #include <vector>
    using namespace std;
    struct Mouse {
    	int Order;
    	int Weight;
    }Mice[1001];
    queue<int> MiceQueue;   //按照队列排序
    bool weightCmp(int a, int b) {
    	return Mice[a].Weight > Mice[b].Weight;
    }
    int main() {
    	int NP, NG;
    	scanf("%d %d", &NP, &NG);
    	for (int i = 0; i < NP; ++i) {
    		scanf("%d", &Mice[i].Weight);
    	}
    	for (int i = 0; i < NP; ++i) {
    		int order;
    		scanf("%d", &order);
    		MiceQueue.push(order);
    	}
    	int temp = NP;  
    	int group = temp % NG == 0 ? temp / NG : temp / NG + 1;
    	while (!MiceQueue.empty()) {
    		for (int i = 1; i <= group; ++i) {
    			int k = NG;
    			if (i == group) k = temp % NG == 0 ? NG : temp % NG;
    			int MAX = 0;
    			int MAXSerise;
    			for (int j = 0; j < k; ++j) {
    				if (Mice[MiceQueue.front()].Weight > MAX) {
    					MAX = Mice[MiceQueue.front()].Weight;
    					MAXSerise = MiceQueue.front();
    				}
    				Mice[MiceQueue.front()].Order = group + 1;  //每轮淘汰的老鼠把排名设置为group+1
    				MiceQueue.pop();
    			}
    			MiceQueue.push(MAXSerise);
    		}
    		if (MiceQueue.size() == 1) {  //最后一只老鼠就是第一名
    			Mice[MiceQueue.front()].Order = 1;
    			MiceQueue.pop();
    		}
    		temp = group;
    		group = temp % NG == 0 ? temp / NG : temp / NG + 1;  //得到下一轮的组数
    	}
    	for (int i = 0; i < NP; ++i) {
    		printf("%d", Mice[i].Order);
    		if (i < NP - 1) printf(" ");
    	}
    
    	system("PAUSE");
    	return 0;
    }

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