CodeForces 660D Number of Parallelograms

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<string>
#include<memory.h>
#include<functional>
#include<limits.h>
#include<vector>
#include<algorithm>
#include<cmath>
#include<iomanip>
#include<set>
#include<map>
using namespace std;
const int maxn = 2005;
struct point {
int x, y;
point(int x1 = 0, int y1 = 0) :x(x1), y(y1) {}
}p[maxn];

struct vec {
int x, y;
vec(int x1, int y1) :x(x1), y(y1) {}
bool operator==(const vec& v) const {
return x == v.x && y == v.y;
}
bool operator<(const vec& v) const {
if (x == v.x)
return y < v.y;
return x < v.x;
}
};

int main() {
int n;
while (cin >> n) {
for (int i = 0; i < n; ++i) {
cin >> p[i].x >> p[i].y;
}
map<vec, int> mp;

for (int i = 0; i < n; ++i) {
for (int j = i + 1; j < n; ++j) {
int x = p[i].x - p[j].x;
int y = p[i].y - p[j].y;
if (x < 0)
x = -x, y = -y;
if (x == 0 && y < 0)
y = -y;
vec temp1(x, y);
mp[temp1]++;
}
}
bool ok = false;
long long cnt = 1;
map<vec, int>::iterator it;
for (it = mp.begin(); it != mp.end(); ++it) {
if (it->second >= 2) {
ok = true;
cnt += it->second * (it->second-1) / 2;
}
}
if (ok)
cout << cnt/2 << endl;
else
cout << 0 << endl;
}

return 0;
}

暴力枚举，找到一组相等的向量→肯定可以构成一个平行四边形（肯定还会出现它的另一对边，这里相当于多算了一次，故最后结果除以2）