PAT

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:

11 325 18 0 46 37 3 19 22 57 56 106 0 8 7 10 5 9 1 4 2 3

Sample Output:

5 5 5 2 5 5 5 3 1 3 5

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给定条件：
1.n只鼠，m只一组
2.n只鼠的体重
3.n只鼠的顺序
4.排序要求：

1.每m只一组，这m只中的最大的获胜，并会进行下一场比赛
2.如果最后不够m只，那么剩下的单独成一组
3.直至剩下最后一个为第一名
4.同时失败的排名相同
5.解释说明：例如11只鼠3只一组能分成4组，最后一组为2只，这次比赛后失败的人名次为5，因为4组分别出一个，占据了第1，2，3，4名，那么失败的就都是第5名

要求：
1.输出排名信息

求解：
1.首先输入数据，按照要求顺序将鼠入队列
2.最直接想法，我们可以遍历队列，把每组最大的元素添加到新队列即可，但是这样会涉及到多个队列，操作复杂
3.我们只要将操作控制在原队列元素个数之内，我们就可以直接将最大的元素直接加到队列的队尾
4.每次操作完成，更新原队列元素个数
5.接着进行一系列入队出队操作，模拟比较进行，直至队列中只剩下一只鼠即可

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#include <cstdio>#include <queue>using namespace std;struct Mice {int id;int weight;int rank;} mice[1005];int np, ng;int order;queue<Mice> q;int main() {scanf("%d%d", &np, &ng);for(int i = 0; i < np; i++) {scanf("%d", &mice[i].weight);mice[i].id = i;}for(int i = 0; i < np; i++) {scanf("%d", &order);q.push(mice[order]);}while(!q.empty()) {int qSize = q.size();int groups = qSize / ng;if(qSize % ng) groups ++;if(qSize == 1) {int id = q.front().id;mice[id].rank = 1;break;}Mice temp;temp.weight = -1;for(int i = 1; i <= qSize; i++) {Mice cur = q.front();int id = cur.id;mice[id].rank = groups + 1;if(temp.weight < cur.weight) { // 找到fattesttemp.id = cur.id;temp.weight = cur.weight;}if(i % ng == 0 || i == qSize) { q.push(temp);temp.weight = -1;}q.pop();}}for(int i = 0; i < np; i++) {if(i) printf(" ");printf("%d", mice[i].rank);}printf("\n");return 0;}

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