PAT Advanced1056 Mice and Rice（队列，模拟）_综合

链接：PAT Advanced1056

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 87 10 5 9 1 4 2 3

Sample Output:

5 5 5 2 5 5 5 3 1 3 5

题意：
给出 N_P 只老鼠的质量，并给出它们的初始顺序，按这个初始顺序把这这些老鼠按每 N_G 只分为一组，最后不够 N_G 只的也单独分为一组。对每组老鼠，选出它们中质量最大的 1 只晋级，这样晋级的老鼠数就等于该轮分组的组数。对这些晋级的老鼠再按上面的步骤每 N_G 只分为一组进行比较，选出质量最大的一批继续晋级，这样到最后只剩下一只老鼠，排名为1。把这些老鼠的排名按原输入的顺序愉出。

注意， 同一轮被淘汰的老鼠名次相同，其名次取决于有多少老鼠排在他们前面，如样例中，第一轮4只老鼠晋级，7只老鼠淘汰，那么这7只老鼠的名次均为5。

方法：
将老鼠的序号以初始顺序存入队列中，每次按规则处理同时让晋级的老鼠重新入队列，当队列中只剩一只老鼠时结束。

以下代码：

#include<iostream>
#include<queue>
#include<cmath>
#include<algorithm>
using namespace std;
struct mouse
{
    int w;int rank;
}a[1010];
int main()
{
    int i,NP,NG;queue<int> q;cin>>NP>>NG;for(i=0;i<NP;i++)cin>>a[i].w;for(i=0;i<NP;i++){
    int t;cin>>t;q.push(t);}while(q.size()>1){
    int left=q.size(),maxw=0,maxi;    //left表示该轮还有多少老鼠for(i=1;i<=left;i++){
    a[q.front()].rank=ceil(left*1.0/NG)+1;   //算出该轮被淘汰老鼠的名次if(a[q.front()].w>maxw)                  //即晋级的老鼠数+1{
                                            //晋级的老鼠数=left/NG向上取整maxw=a[q.front()].w;maxi=q.front();}if(i%NG==0||i==left)          //每NG个一组{
    q.push(maxi);maxw=0;}q.pop();}}a[q.front()].rank=1;    //最后的剩下的一个排名为 1for(i=0;i<NP;i++){
    if(i!=0)cout<<" ";cout<<a[i].rank;}return 0;
}