1056. Mice and Rice (25)

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<cstdio>
#include<queue>
using namespace std;
const int maxn = 1010;
struct mouse{
    int weight;
    int R; //排名 
}mouse[maxn];

int main(){
    int np,ng;//比赛人数，分组人数 
    scanf("%d%d",&np,&ng);
    int i,order;
    for(i = 0; i < np; i++){
        scanf("%d",&mouse[i].weight);
    }
    queue<int> q;
    for(i = 0; i < np; i++){
        scanf("%d",&order);
        q.push(order);
    }
    int temp = np,group; //temp为每轮比较人数，第一轮为np个 
    while(q.size() != 1){
        if(temp % ng == 0) group = temp/ng; //比赛人数是否可以整分 
        else group = temp/ng + 1;
        for(i = 0; i < group; i++){
            int k = q.front();
            for(int j = 0; j < ng; j++){
                if(i * ng + j >= temp) break; //= 取到是因为j从零开始取的  退出最后一组小组比较循环条件 
                int front = q.front();
                if(mouse[front].weight > mouse[k].weight) k = front;
                mouse[front].R = group + 1;
                q.pop();
            }
            q.push(k);
        }
        temp = group;
    }
    mouse[q.front()].R = 1;
    for(i = 0; i < np; i++){
        printf("%d",mouse[i].R);
        if(i < np - 1) printf(" ");
        else printf("\n");
    }

    return 0;
}