PTA Mice and Rice (25分)

释放无限光明的是人心，制造无边黑暗的也是人心，光明和黑暗交织着，厮杀着，这就是我们为之眷恋又万般无奈的人世间。

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 <iostream>
#include <cstdio>
#include <algorithm>
#include <string>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <vector>
#include <ctime>
#include <cctype>
#include <bitset>
#include <utility>
#include <sstream>
#include <complex>
#include <iomanip>
#include<climits>//INT_MAX
//#include<bits/stdc++.h>
#define PP pair<ll,int>
#define inf 0x3f3f3f3f
#define llinf 0x3f3f3f3f3f3f3f3fll
#define dinf 1000000000000.0
#define PI 3.1415926
typedef long long ll;
using namespace std;
int const mod=1e9+7;
const int maxn=1e4+10;
struct node{
    int w;
    int jg;
}P[maxn];
int np,ng;
int main()
{
    scanf("%d%d",&np,&ng);
    for(int i=0;i<np;i++)
        scanf("%d",&P[i].w);
    queue<int>q;
    for(int i=0;i<np;i++)
    {
        int x;
        scanf("%d",&x);
        q.push(x);
    }
    int tmp=np,d;
    while(q.size()!=1)
    {
        if(tmp%ng==0)
            d=tmp/ng;
        else
            d=tmp/ng+1;
        for(int i=0;i<d;i++)
        {
            int k=q.front();
            for(int j=0;j<ng;j++)
            {
                if(i*ng+j>=tmp)
                    break;
                int v=q.front();
                q.pop();
                if(P[v].w>P[k].w)
                    k=v;
                P[v].jg=d+1;
            }
            q.push(k);
        }
        tmp=d;
    }
    P[q.front()].jg=1;
    for(int i=0;i<np;i++)
    {
        printf("%d",P[i].jg);
        if(i!=np-1)
            printf(" ");
    }
    return 0;
}