PAT甲级1056

1056. Mice and Rice (25)

时间限制

30 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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<queue>
#include<vector>
#include<algorithm>
using namespace std;

struct person
{
  int id, rank,weight;
  bool operator<(person p)
  {
    return rank < p.rank;
  }
};
int cmp(person p1, person p2)
{
  return p1.id < p2.id;
}
int main()
{
  int Np, Ng;
  cin >> Np >> Ng;
  queue<person> q;
  vector<person> v;
  int t;
  person p;
  for (int i = 0; i < Np; i++)
  {
    cin >> t;
    p.id = i;
    p.weight = t;
    v.push_back(p);
  }
  
  for (int i = 0; i < Np; i++)
  {
    cin >> t;
    p.id = t;
    p.weight = v[t].weight;
    q.push(p);
  }
  int winnerid, maxweight = -1;
  int count, size=1001;
  int relativerank = 1001;
  while (size>1)
  {
    size = q.size();
    count = 0;
    while (count < size)
    {
      if (maxweight < q.front().weight)
      {
        winnerid = q.front().id;
        maxweight = q.front().weight;
      }
      v[q.front().id].rank = relativerank;
      q.pop();
      count++;
      if (count % Ng == 0|| count == size)
      {
        p.id = winnerid;
        p.weight = maxweight;
        q.push(p);
        maxweight = -1;
      }
    }
    relativerank--;  
  }
  sort(v.begin(), v.end());
  int oldrank = 1, oldrelativerank;
  for (int i = 0; i < Np; i++)
  {
    if (!i)
    {
      oldrelativerank = v[i].rank;
      v[i].rank = oldrank;
    }
    else
    {
      if (v[i].rank == oldrelativerank)
        v[i].rank = oldrank;
      else
      {
        oldrank = i + 1;
        oldrelativerank = v[i].rank;
        v[i].rank = oldrank;
      }
    }
  }
  sort(v.begin(), v.end(), cmp);
  for (int i = 0; i < Np; i++)
  {
    if (!i)
      cout << v[i].rank;
    else
      cout << " " << v[i].rank;
  }
  return 0;
}
/*做复杂了，晋级的顺序没分析清楚。其实可以根据第一次晋级的老鼠数目来确定最后一名是多少*/