이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include<bits/stdc++.h>
using namespace std;
#define int long long
#define ii pair <int, int>
#define app push_back
#define all(a) a.begin(), a.end()
#define bp __builtin_popcountll
#define ll long long
#define mp make_pair
#define f first
#define s second
#define Time (double)clock()/CLOCKS_PER_SEC
const int N = 3007;
struct Point {
int x, y, c;
Point operator - (Point p) {
return {x - p.x, y - p.y};
}
int operator *(Point p) {
return x * p.y - y * p.x;
}
} a[N];
int sign(int x) {
if (x < 0)
return -1;
else if (x == 0)
return 0;
else
return 1;
}
int ans[N][N];
signed main() {
#ifdef HOME
freopen("input.txt", "r", stdin);
#else
#define endl '\n'
ios_base::sync_with_stdio(0); cin.tie(0);
#endif
int n, m;
cin >> n >> m;
for (int i = 1; i <= n; ++i) {
cin >> a[i].x >> a[i].y >> a[i].c;
}
Point A, B;
cin >> A.x >> A.y;
cin >> B.x >> B.y;
for (int i = 1; i <= n; ++i) {
auto l = A - a[i];
auto r = B - a[i];
for (int j = 1; j <= n; ++j) {
if (a[i].c == a[j].c)
continue;
auto v = a[j] - a[i];
if (sign(l * r) * sign(l * v) > 0 && sign(l * r) * sign(v * r) > 0) {
++ans[a[i].c][a[j].c];
}
}
}
int q;
cin >> q;
while (q--) {
int c1, c2;
cin >> c1 >> c2;
cout << ans[c1][c2] << endl;
}
}
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