PAT A 1056 Mice and Rice

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

佩服

#include<cstdio>
#include<queue>
using namespace std;

const int maxn = 1010;

struct mouse {
    int wt;
    int r;
}M[maxn];

int main() {
    int np, ng, order;
    scanf("%d %d", &np, &ng);
    queue<int> q;
    for (int i = 0; i < np; i++) {
        scanf("%d", &M[i].wt);
    }

    for (int i = 0; i < np; i++) {
        scanf("%d", &order);
        q.push(order);
    }

    int num = np, group, max;
    while (num != 1) {
        if (num % ng == 0) group = num / ng;
        else group = num / ng + 1;
        for (int i = 0; i < group; i++) {
            max = q.front();
            for (int j = 0; j < ng; j++) {
                if (i*ng + j == num) break;//判断最后一组的数据，不能用i==group-1 && j==num % ng 因为如果num能够整除ng，就出错了。。。
                int temp = q.front();
                if (M[temp].wt > M[max].wt) max = temp;
                M[temp].r = group + 1;
                q.pop();
            }
            q.push(max);
        }
        num = group;
    }
    M[q.front()].r = 1;//这里要写q.front(),否则对于np=1，会出错。

    for (int i = 0; i < np; i++) {
        printf("%d", M[i].r);
        if (i < np - 1) printf(" ");
    }
    return 0;
}