PAT1056:Mice and Rice

1056. Mice and Rice (25)

时间限制

100 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

思路
  有点绕的逻辑划分题，注意第三行的数据是选手的编号，索引就是选手的出场顺序,另外当前失败选手的rank = 晋级的选手数+1
代码

#include<vector>
#include<algorithm>
#include<iostream>
#include<queue>
using namespace std;
class rat
{
  public:
      int weight;
      int rank;
      int id;
      int order;
};


bool cmp(rat& a,rat& b)
{
    return a.id < b.id;
}
int main()
{
    int Np,Ng;
    vector<rat> rats;
    while(cin >> Np >> Ng)
    {
       vector<int> weights(Np);
       rats.resize(Np);
       for(int i = 0;i < Np;i++)
       {
           cin >> weights[i];
       }
       for(int i = 0;i < Np;i++)
       {
           cin >> rats[i].id ;
           rats[i].order = i;
           rats[i].weight = weights[rats[i].id];
       }
       queue<rat> q;
       for(int i = 0;i < Np;i++)
           q.push(rats[i]);
       while(!q.empty())
       {
           int s = q.size();
           if(s == 1)
           {
               rat tmp = q.front();
               rats[tmp.order].rank = 1;
               break;
           }
           int levels = s / Ng + (s % Ng == 0?0:1);
           rat maxrat;
           int maxn = -1,cnt = 0;
           for(int i = 0;i < s;i++)
           {
               rat tmp = q.front();
               rats[tmp.order].rank = levels + 1;
               q.pop();cnt++;
               if(tmp.weight > maxn)
               {
                   maxn = tmp.weight;
                   maxrat = tmp;
               }
               if(cnt == Ng || i == s - 1)
               {
                   cnt = 0;
                   maxn = -1;
                   q.push(maxrat);
               }
           }
       }
        sort(rats.begin(),rats.end(),cmp);
           for(int i = 0;i < Np;i++)
           {
               if(i != 0)
                cout << " ";
               cout << rats[i].rank;
           }
    }
}