Created
October 23, 2011 16:44
-
-
Save FooBarrior/1307569 to your computer and use it in GitHub Desktop.
south-2011:C. Doesn't work.
This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. To review, open the file in an editor that reveals hidden Unicode characters.
Learn more about bidirectional Unicode characters
#include <iostream> | |
#include <cstdlib> | |
#include <vector> | |
#include <algorithm> | |
#include <cstdio> | |
using namespace std; | |
struct Line{ | |
int x1, y1, x2, y2; | |
bool operator<(const Line &l){ | |
return x1 == x2 ? y1 < l.y1 : x1 < l.x1; | |
} | |
Line(istream& is){ | |
is >> x1 >> y1 >> x2 >> y2; | |
if(x1 > x2) swap(x1, x2); | |
if(y1 > y2) swap(y1, y2); | |
} | |
}; | |
struct Pt{ | |
int h, v, x, y; | |
int hpos, vpos; | |
Pt(int h, int v, int hp, int vp, int x, int y):h(h), v(v),hpos(hp), vpos(vp), x(x), y(y){} | |
friend ostream& operator<<(ostream &os, Pt &p){ | |
return os << p.x << ":" << p.y << " "; | |
} | |
}; | |
int main(){ | |
#ifndef ONLINE_JUDGE | |
freopen("in.txt", "r", stdin); | |
#endif | |
int n; | |
cin >> n; | |
vector<Line> h, v; | |
for(int i = 0; i < n; i++){ | |
Line l(cin); | |
(l.y1 == l.y2 ? h : v).push_back(l); | |
} | |
sort(h.begin(), h.end()); | |
sort(v.begin(), v.end()); | |
vector<vector<Pt> > hp(h.size()), vp(v.size()); | |
for(int i = 0; i < h.size(); i++){ | |
for(int j = 0; j < v.size(); j++){ | |
Line &hl = h[i], &vl = v[j]; | |
if(hl.x1 <= vl.x1 && hl.x2 >= vl.x2 && vl.y1 <= hl.y1 && vl.y2 >= hl.y2){ | |
Pt p(i, j, hp[i].size(), vp[j].size(), vl.x1, hl.y1); | |
hp[i].push_back(p); | |
vp[j].push_back(p); | |
} | |
} | |
} | |
int cnt = 0; | |
for(int i = 0; i < hp.size(); i++){ | |
vector<Pt> &hh = hp[i]; | |
for(int j = 0; j < hh.size(); j++){ | |
Pt &p = hh[j]; | |
for(int k = p.hpos + 1; k < hh.size(); k++){ | |
Pt &ph1 = hh[k]; | |
vector<Pt> &vv = vp[ph1.v]; | |
for(int l = ph1.vpos + 1; l < vv.size(); l++){ | |
Pt &pv1 = vv[l]; | |
vector<Pt> &hh2 = hp[pv1.h]; | |
for(int m = 0; m < pv1.hpos; m++){ | |
Pt &ph2 = hh[m]; | |
if(ph2.v == p.v){ | |
cnt++; | |
//cout << p << ph1 << pv1 << ph2 << endl; | |
break; | |
} | |
} | |
} | |
} | |
} | |
} | |
cout << cnt; | |
return 0; | |
} |
Sign up for free
to join this conversation on GitHub.
Already have an account?
Sign in to comment
Im fool. solution is http://codeforces.ru/blog/entry/2923#comment-59324