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.
老鼠和大米是一个程序竞赛的名字，这个竞赛里的程序需要写代码来控制老鼠在给定地图上的移动。每一个老鼠的目标是吃尽可能多的大米来变成更胖的老鼠。
最开始NP个编程者游戏顺序是随机决定的，每NG个编程者在比赛中为一组。组内最胖的老鼠胜利并进入下一轮比赛。这一轮中所有的失败者排名相同。每NG个获胜者为一组进入下一个比赛。
为了简单起见，假设一旦程序员提交了他/她的代码，每个鼠标的重量都是固定的。给出所有鼠标的重量和初始播放顺序，您应该为程序员输出排名。
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 NGmice 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.
第一行包含2个正数：NP和NG。第二行包含NP个互不相同的非负数Wi 分别表示第i个老鼠的重量。第三行给出初始顺序
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.
给出最终排名 第i个数字是第i个程序员的排名。
#include<cstdio>
#include<queue>
using namespace std;
const int maxn=1010;
struct mouse{
int weight;//质量
int R;//排名
}mouse[maxn];
int main(){
int np,ng,order;
scanf("%d %d",&np,&ng);
for(int i=0;i<np;i++){
scanf("%d",&mouse[i].weight);
}
queue<int> q;
for(int i=0;i<np;i++){
scanf("%d",&order);
q.push(order);
}
int temp=np,group;
while(q.size()!=1){
//分组
if(temp%ng==0) group = temp/ng;
else group = temp/ng +1;
//遍历每一组 用k存储每一组胜利者编号
for(int i=0;i<group;i++){
int k = q.front();
for(int j=0;j<ng;j++){
if(i*ng+j>=temp) break;//超出np，就停止循环
int front = q.front();
if(mouse[front].weight>mouse[k].weight)
k=front;
mouse[front].R = group+1;
q.pop();//出队
}
q.push(k);//晋级者入队
}
temp=group;//group只老鼠晋级，因此下一轮老鼠数为group
}
//当队列中只剩1只老鼠时，令其排名为1
mouse[q.front()].R=1;
for(int i=0;i<np;i++){
printf("%d",mouse[i].R);
if(i<np-1) printf(" ");
}
return 0;
} 