PAT A1056

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

#include "stdio.h"
//#include "math.h"
#include "string.h"
//#include "iostream"
#include "stdlib.h"
//#include "vector"
//#include "set"
//#include "map"
//#include "stack"
#include "queue"
#include "algorithm"
using namespace std;

//typedef long long LL;
const int maxn = 1010;
struct mouse{
    //老鼠
    int weight;//质量
    int R;//排名
}mouse[maxn];

int main(){
    int np, ng, order;
    scanf("%d%d", &np, &ng);//number of problem 、 number of group
    for (int i = 0; i < np; i++) {
        scanf("%d", &mouse[i].weight);
    }
    queue<int>  q;//定义一个队列
    for (int i = 0; i < np; i++) {
        scanf("%d", &order);//题目给出的顺序
        q.push(order);//按顺序把老鼠们的标号入队
    }
    int temp = np, group;//temp为当前轮的比赛总老鼠数，group为组数
    while (q.size() != 1) {
        //计算group， 即当前轮分为几组进行比赛
        if (temp % ng == 0) group = temp / ng;
        else group = temp / ng + 1;
        //枚举每一组，选出该组中老鼠质量最大的
        for (int i = 0; i < group ; i++) {
            int k = q.front();//k存放该组质量最大的老鼠的编号
            for (int j = 0; j < ng; j++){
                //在最后一组老鼠数不足NG时起作用，退出循环
                if (i * ng + j >= temp) {
                    break;
                }
                int front = q.front();
                if (mouse[front].weight > mouse[k].weight) {
                    k = front;//找出质量最大的老鼠
                }
                mouse[front].R = group + 1;//该轮老鼠的排名为group+1
                q.pop();
            }
            q.push(k);
        }
        temp = group; //group只老鼠晋级，因此下轮总老鼠数为gruop
    }
    mouse[q.front()].R = 1;//当队列中只剩1只老鼠时，令其排名为1
    //输出所有老鼠的信息
    for (int i = 0; i < np; i++) {
        printf("%d", mouse[i].R);
        if (i < np - 1) {
            printf(" ");
        }
    }
    return 0;
}

原文地址：https://www.cnblogs.com/sanxiandoupi/p/11692413.html