1056. Mice and Rice (25)*

最新推荐文章于 2025-03-15 10:48:16 发布

原创最新推荐文章于 2025-03-15 10:48:16 发布 · 515 阅读

0 ·

CC 4.0 BY-SA版权

PAT 专栏收录该内容

79 篇文章

订阅专栏

Mice and Rice 是一场编程比赛，参赛者需编写代码控制小鼠在地图上的行动，目标是吃尽可能多的米成为肥鼠。比赛首先随机决定参赛者的比赛顺序，每 N 个参赛者组成一组进行匹配。每组中体重最大的小鼠获胜进入下一轮，其余参赛者在同一轮排名相同。直至决出最后的胜者。

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

//#include "stdafx.h"
#include <cstdio>
#include <cstring>

const int N=1005;
int mouse[N],order[N],rank[N];
bool used[N];

int main()
{
	//freopen("D://CodeC++/PAT_1056/1.txt","r",stdin);
    int NP,NG,i,group;
	bool flag=true;
	memset(used,0,sizeof(used));
	scanf("%d %d",&NP,&NG);
	for(i=0;i<NP;++i)
		scanf("%d",mouse+i);
	for(i=0;i<NP;++i)
		scanf("%d",order+i);
	NP%NG==0?group=NP/NG:group=NP/NG+1;

	int last,max,countNG;
	while(flag){
		if(group==1)
			flag=false;
		countNG=0,last=-1,max=-1;

		for(i=0;i<NP;++i){
			if(!used[order[i]]){
				if(countNG==NG){
					last=-1;
					max=-1;
					countNG=1;
				}
				else
					++countNG;
				if(mouse[order[i]]>max){
					if(last!=-1){
						rank[last]=group+1;
						used[last]=true;
					}
					last=order[i];
					max=mouse[order[i]];
				}
				else{
					rank[order[i]]=group+1;
					used[order[i]]=true;
				}
			}
		}
		group%NG==0?group=group/NG:group=group/NG+1;
	}
	rank[last]=1;

	for(i=0;i<NP;++i)
		i==NP-1?printf("%d\n",rank[i]):printf("%d ",rank[i]);

    return 0;
}