A1062 Talent and Virtue (25 points | sort, with detailed comments, logic analysis)

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EDITORIAL

• Ideas analysis
  • Student structured array package meta information
    • An array of structures vector v [4] category candidates package 4
  • cmp function, time-consuming than
    • Sort according to first sort out, and then follow sorted by Germany, according to the last sort Caifen
  • Output meets criteria -
• Simple title, 20 minutes a title

Test Case

• input:
  14 60 80
  10000001 64 90
  10000002 90 60
  10000011 85 80
  10000003 85 80
  10000004 80 85
  10000005 82 77
  10000006 83 76
  10000007 90 78
  10000008 75 79
  10000009 59 90
  10000010 88 45
  10000012 80 100
  10000013 90 99
  10000014 66 60
  
  output:
  12
  10000013 90 99
  10000012 80 100
  10000003 85 80
  10000011 85 80
  10000004 80 85
  10000007 90 78
  10000006 83 76
  10000005 82 77
  10000002 90 60
  10000014 66 60
  10000008 75 79
  10000001 64 90
  

ac Code

• #include <iostream>
  #include <algorithm>
  #include <vector>
  using namespace std;
  struct node
  {
      int num, de, cai;
  };
  
  int cmp(node a,node b)
  {
      if((a.de + a.cai) != (b.de+b.cai))
          return (a.de + a.cai) > (b.de+b.cai);
      else if(a.de != b.de)
          return a.de>b.de;
      else
          return a.num<b.num;
  }
  
  int main()
  {
      int n, low, high;
      scanf("%d %d %d", &n, &low, &high);
      vector<node> v[4];
      node tmp;
      int total = n;
  
      for(int i=0; i<n; i++)
      {
          scanf("%d %d %d", &tmp.num,&tmp.de,&tmp.cai);
          if(tmp.de < low || tmp.cai < low)
              total--;
          else if(tmp.de >= high && tmp.cai >= high)
              v[0].push_back(tmp);
          else if(tmp.de >= high && tmp.cai<high)
              v[1].push_back(tmp);
          else if(tmp.de < high && tmp.cai < high && tmp.de >= tmp.cai)
              v[2].push_back(tmp);
          else
              v[3].push_back(tmp);
      }
  
      printf("%d\n", total);
      for(int i=0; i<4; i++)
      {
          sort(v[i].begin(), v[i].end(), cmp);
          for(int j=0; j<v[i].size(); j++)
              printf("%d %d %d\n", v[i][j].num, v[i][j].de, v[i][j].cai);
      }
  
      return 0;
  }