270 - Lining Up/270.2.cpp

   1 /*
   2   Problem: 270 - Lining up
   3   Andrés Mejía-Posada (andmej@gmail.com)
   4
   5 */
   6 using namespace std;
   7 #include <algorithm>
   8 #include <iostream>
   9 #include <iterator>
  10 #include <sstream>
  11 #include <fstream>
  12 #include <cassert>
  13 #include <climits>
  14 #include <cstdlib>
  15 #include <cstring>
  16 #include <string>
  17 #include <cstdio>
  18 #include <vector>
  19 #include <cmath>
  20 #include <queue>
  21 #include <deque>
  22 #include <stack>
  23 #include <list>
  24 #include <map>
  25 #include <set>
  26
  27 #define D(x) cout << #x " is " << x << endl
  28
  29 struct point{
  30   int x, y;
  31   point(int x=0, int y=0) : x(x), y(y) {}
  32 };
  33
  34 void simplify(pair<int, int> &p){
  35   #define x first
  36   #define y second
  37   assert(p.x != 0 || p.y != 0);
  38   if (p.y == 0){
  39     p.x = 1;
  40     return;
  41   }
  42   if (p.x == 0){
  43     p.y = 1;
  44     return;
  45   }
  46   int sign = ((p.x<0)^(p.y<0) ? -1 : 1);
  47   p.x = abs(p.x), p.y = abs(p.y);
  48   int g = __gcd(p.x, p.y);
  49   if (g) p.x /= g, p.y /= g;
  50   p.x *= sign;
  51   #undef x
  52   #undef y
  53 }
  54
  55 int main(){
  56   int casos;
  57   cin >> casos;
  58   string s;
  59   getline(cin, s), getline(cin, s);
  60   while (casos--){
  61     vector<point> p;
  62     while (getline(cin, s) && s != ""){
  63       point _;
  64       sscanf(s.c_str(), "%d %d", &_.x, &_.y);
  65       p.push_back(_);
  66     }
  67
  68     int x = 0;
  69     map<pair<int, int>, int> cnt;
  70     const int n = p.size();
  71     for (int i=0; i<n; ++i){
  72       for (int j=i+1; j<n; ++j){
  73         pair<int, int> slope(p[j].y - p[i].y, p[j].x - p[i].x);
  74         simplify(slope);
  75         map<pair<int, int>, int>::iterator i = cnt.find(slope);
  76         if (i != cnt.end()) x = max(x, ++i->second);
  77         else x = max(x, cnt[slope] = 2);
  78       }
  79     }
  80     cout << (int)((1 + sqrt(1 + 8*x))/2) << endl;
  81     if (casos > 0) cout << endl;
  82   }
  83   return 0;
  84 }