PAT Advanced 1056 Mice and Rice (25) [queue的⽤法]

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 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 lef 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

题目分析

已知参赛人数N，每组人数M，参赛者体重，起始参赛顺序。每组最重者获胜，每轮有M个人获胜进入下一组进行下一轮比赛

注：因为每轮比赛，有M个人获胜，所以其余失败者排名为M+1，获胜者排名为M

解题思路

1. 定义queue队列，模拟比赛流程，每场比赛所有参赛者入队，并在最后设置哨兵（哨兵表明本轮比赛结束）
2. 与层级遍历树结构的差别：本题目中退出队列循环标识为参赛人数为1，而层级遍历树退出队列循环标识为队列中没有节点
3. 每次循环到下一组第一个参赛者时，将前一组的获胜者设置排名添加到队列最后，准备下一场比赛

Code

#include <iostream>
#include <queue>
using namespace std;
struct M {
	int w=-1;//weight
	int i=-1;//id
	int r=-1;//rank
};
int main(int argc,char * argv[]) {
	int np,ng,t;
	scanf("%d%d",&np,&ng);
	M ms[np+1];
	for(int i=0; i<np; i++) {
		scanf("%d",&ms[i].w);
		ms[i].i=i;
	}
	M em;
	ms[np]=em;
	queue<int> q;
	for(int i=0; i<np; i++) {
		scanf("%d",&t);
		q.push(t);
	}
	int group = np%ng==0?np/ng:np/ng+1;
	int max=np,cs=np;
	q.push(-1);//第一轮最后加一个哨兵
	for(int i=1; group>=1&&q.size()>=1; i++) {
		int cur = q.front();
		if(cur!=-1)ms[cur].r = group+1;
		if(cur==-1) { // 若本轮比赛结束
			q.pop(); //将哨兵弹出，方便计算下一场比赛人数
			ms[max].r = group; //添加上一轮比赛最后一组的获胜者到队列，预备下一场比赛
			if(cs==1)break; //如果当前参赛人员只有一人，并且已完成排名，退出
			q.push(max); //将本组获胜者加入队列，预备下一轮比赛
			cs=q.size(); //下一轮参赛人数
			group=(cs%ng==0?cs/ng:cs/ng+1); //重新计算group
			q.push(-1); //每一轮最后加一个哨兵
			max=np; //max置为哨兵参照，weight=-1
			i=0; //重置 指针
			continue;
		} else if(i!=1&&i%ng==1) { //一组统计完成，当前为下一组第一个参赛者
			ms[max].r = group;
			q.push(max);//将本组获胜者加入队列，预备下一轮比赛
			max=cur; //cur为下一组的第一个参赛者
		} else if(ms[max].w<ms[cur].w) { //更新本组获胜者
			max=cur;
		}
		q.pop();
	}
	for(int i=0; i<np; i++) {
		if(i!=0)printf(" ");
		printf("%d",ms[i].r);
	}
	return 0;
}