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 }

 

转载于:https://www.cnblogs.com/xiaoyesoso/p/4282386.html