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

  1 #include<stdio.h>

  2 #include<vector>

  3 #include<map>

  4 #include<algorithm>

  5 using namespace std;

  6 

  7 struct MyStruct

  8 {

  9     int ID;

 10     int wight;

 11 };

 12 

 13 struct MyRank

 14 {

 15     int time;

 16     int ID;

 17 };

 18 

 19 vector<MyRank> ranking;

 20 

 21 void fun(vector<MyStruct> vv ,int Ng)

 22 {

 23     vector<MyStruct> tem;

 24     for(int i = 0;i< vv.size(); i++)

 25     {

 26         if( i + Ng -1 < vv.size())

 27         {

 28             int MAX = -1;

 29             int index;

 30             int j;

 31             for(j = i ;j < i + Ng ;j++)

 32             {

 33                 if(vv[j].wight > MAX)

 34                 {

 35                     MAX = vv[j].wight;

 36                     index = j;

 37                 }

 38             }

 39             i = j-1;

 40             tem.push_back(vv[index]);

 41             ++ranking[vv[index].ID].time;

 42         }

 43         else

 44         {

 45             int MAX = -1;

 46             int index;

 47             int j;

 48             for(j = i ;j < vv.size() ;j++)

 49             {

 50                 if(vv[j].wight > MAX)

 51                 {

 52                     MAX = vv[j].wight;

 53                     index = j;

 54                 }

 55             }

 56             i = j;

 57             tem.push_back(vv[index]);

 58             ++ranking[vv[index].ID].time;

 59         }

 60     }

 61 

 62     if(tem.size() > 1)

 63         fun(tem,Ng);

 64 }

 65 

 66 int cmp(MyRank a,MyRank b)

 67 {

 68     return a.time > b.time ;

 69 }

 70 

 71 int main()

 72 {

 73     int Np,Ng,i;

 74     MyStruct tem;

 75     MyRank Rtem;

 76     int index;

 77     scanf("%d%d",&Np,&Ng);

 78     vector<MyStruct> v1,v2;

 79     for(i = 0;i< Np ;i++)

 80     {

 81         scanf("%d",&tem.wight);

 82         tem.ID = i;

 83         Rtem.ID = i;

 84         Rtem.time = 0;

 85         ranking.push_back(Rtem);

 86         v1.push_back(tem);

 87     }

 88     for(i = 0;i< Np ;i++)

 89     {

 90         scanf("%d",&index);

 91         v2.push_back(v1[index]);

 92     }

 93 

 94     fun(v2,Ng);

 95 

 96     sort(ranking.begin(),ranking.end(),cmp);

 97     int result[1001];

 98     int j = 1;

 99     for(i = 0 ;i <ranking.size();i++)

100     {

101         result[ranking[i].ID] = j;

102         int count = 1;

103         int tem = ranking[i].time;

104         while(i+1 <ranking.size()&&tem == ranking[i+1].time)

105         {

106             ++i;

107             ++ count;

108             result[ranking[i].ID] = j;

109         }

110         j = j+count;

111     }

112 

113     for(i = 0 ;i < Np ;i++)

114     {

115         if(i == 0) printf("%d",result[i]);

116         else printf(" %d",result[i]);

117     }

118     printf("\n");

119     return 0;

120 }