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_Gwinners 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_iis 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<stdio.h>
 #include<vector>
 #include<map>
 #include<algorithm>
 using namespace std;

 struct MyStruct
 {
     int ID;
     int wight;
 };

 struct MyRank
 {
     int time;
     int ID;
 };

 vector<MyRank> ranking;

 void fun(vector<MyStruct> vv ,int Ng)
 {
     vector<MyStruct> tem;
     for(int i = ;i< vv.size(); i++)
     {
         if( i + Ng - < vv.size())
         {
             int MAX = -;
             int index;
             int j;
             for(j = i ;j < i + Ng ;j++)
             {
                 if(vv[j].wight > MAX)
                 {
                     MAX = vv[j].wight;
                     index = j;
                 }
             }
             i = j-;
             tem.push_back(vv[index]);
             ++ranking[vv[index].ID].time;
         }
         else
         {
             int MAX = -;
             int index;
             int j;
             for(j = i ;j < vv.size() ;j++)
             {
                 if(vv[j].wight > MAX)
                 {
                     MAX = vv[j].wight;
                     index = j;
                 }
             }
             i = j;
             tem.push_back(vv[index]);
             ++ranking[vv[index].ID].time;
         }
     }

     if(tem.size() > )
         fun(tem,Ng);
 }

 int cmp(MyRank a,MyRank b)
 {
     return a.time > b.time ;
 }

 int main()
 {
     int Np,Ng,i;
     MyStruct tem;
     MyRank Rtem;
     int index;
     scanf("%d%d",&Np,&Ng);
     vector<MyStruct> v1,v2;
     for(i = ;i< Np ;i++)
     {
         scanf("%d",&tem.wight);
         tem.ID = i;
         Rtem.ID = i;
         Rtem.time = ;
         ranking.push_back(Rtem);
         v1.push_back(tem);
     }
     for(i = ;i< Np ;i++)
     {
         scanf("%d",&index);
         v2.push_back(v1[index]);
     }

     fun(v2,Ng);

     sort(ranking.begin(),ranking.end(),cmp);
     int result[];
     int j = ;
     for(i =  ;i <ranking.size();i++)
     {
         result[ranking[i].ID] = j;
         int count = ;
         int tem = ranking[i].time;
         while(i+ <ranking.size()&&tem == ranking[i+].time)
         {
             ++i;
             ++ count;
             result[ranking[i].ID] = j;
         }
         j = j+count;
     }

     for(i =  ;i < Np ;i++)
     {
         if(i == ) printf("%d",result[i]);
         else printf(" %d",result[i]);
     }
     printf("\n");
     return ;
 }