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codeForces_13D_D_Triangles.cpp
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77 lines (67 loc) · 1.66 KB
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/**
* 题意：题意，在一个平面上，给定 N 个红色点和 M 个蓝色点（任意三点不共线），求有多少个独立三角形能由
* 红色点组成，且内部不包含蓝色点。
* 限制条件：1. 所有坐标点的横，纵坐标绝对值不大于 10^9
* 2. 0 <= N, M <= 500
* 3. 每个用例运行时间不超过 2s， 运行内存不大于 64M
*
* 解题思路：这是个计算几何的问题，
*
* AC：
* Time: 1122 ms
* Memory:800 KB
*
*/
#include<stdio.h>
#include<map>
#include<utility>
using namespace std;
typedef long long ll;
const int INF = 0x3f3f3f3f, maxn = 505;
struct Point
{
int x, y;
Point() {};
Point(int _x, int _y) {
x = _x;
y = _y;
}
Point operator- (const Point &b) const
{
return Point(x - b . x, y - b . y);
}
ll operator* (const Point &b) const
{
return (ll)x * b . y - (ll)y * b . x;
}
}a[maxn], b[maxn];
int n, m, dp[maxn][maxn];
int main() {
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; i++) {
scanf("%d%d", &a[i] . x, &a[i] . y);
}
for (int i = 1; i <= m; i++) {
scanf("%d%d", &b[i] . x, &b[i] . y);
}
a[0] = Point(-1e9 - 1, -1e9 - 1);
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++) {
if ((a[i] - a[0]) * (a[j] - a[0]) <= 0)
continue;
for (int k = 1; k <= m; k++)
if ((a[i] - a[0]) * (b[k] - a[0]) >= 0 \
&& (a[j] - a[i]) * (b[k] - a[i]) >= 0 \
&& (b[k] - a[0]) * (a[j] - a[0]) >= 0)
dp[i][j]++;
dp[j][i] = -dp[i][j];
}
int ans = 0;
for (int i = 1; i <= n; i++)
for (int j = i + 1; j <= n; j++)
for (int k = j + 1; k <= n; k++)
if (dp[i][j] + dp[k][i] + dp[j][k] == 0)
ans++;
printf("%d\n", ans);
return 0;
}