uva 10256(凸包)

题意：给出n个红点和m个蓝点的坐标，是否存在一条直线，使得取一个红点和一个蓝点都是在直线的异侧。
题解：明显是要判断两个凸包是否有交集，要判断两个凸包的边是否相交，还要判断红点是否在蓝点凸包内部，蓝点是否在红点凸包内部。

#include 
#include 
#include 
#include 
using namespace std;
const double PI = acos(-1);
const int N = 505;
const double INF = 1e9;
struct Point {double x, y;Point(double x = 0, double y = 0): x(x), y(y) {}
}Red[N], Blue[N], resr[N], resb[N];
int n, m;
double Sqr(double x) {return x * x;
}
Point operator + (Point A, Point B) {return Point(A.x + B.x, A.y + B.y);
}
Point operator - (Point A, Point B) {return Point(A.x - B.x, A.y - B.y);
}
Point operator * (Point A, double p) {return Point(A.x * p, A.y * p);
}
Point operator / (Point A, double p) {return Point(A.x / p, A.y / p);
}
//计算点积的正负  负值夹角为钝角
int dcmp(double x) {if (fabs(x) < 1e-9)return 0;return x < 0 ? -1 : 1;
}
bool operator < (const Point& a, const Point& b) {return a.x < b.x || (a.x == b.x && a.y < b.y);
}
bool operator == (const Point& a, const Point& b) {return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0;
}
//计算点积
double Dot(Point A, Point B) {return A.x * B.x + A.y * B.y;
}
//计算叉积，也就是数量积
double Cross(Point A, Point B) {return A.x * B.y - A.y * B.x;
}
//计算向量长度
double Length(Point A) {return sqrt(Dot(A, A));
}
//向量A旋转rad弧度，rad负值为顺时针旋转
Point Rotate(Point A, double rad) {return Point(A.x * cos(rad) - A.y * sin(rad), A.x * sin(rad) + A.y * cos(rad));
}
//角度转化弧度
double torad(double deg) {return deg / 180.0 * PI;
}
//判断两个线段是否有交点(不包括端点)
bool SegmentProperIntersection(Point a1, Point a2, Point b1, Point b2) {double c1 = Cross(a2 - a1, b1 - a1);double c2 = Cross(a2 - a1, b2 - a1);double c3 = Cross(b2 - b1, a1 - b1);double c4 = Cross(b2 - b1, a2 - b1);return dcmp(c1) * dcmp(c2) < 0 && dcmp(c3) * dcmp(c4) < 0;
}
//判断点p是否在线段a1--a2上(不包括端点)
bool OnSegment(Point p, Point a1, Point a2) {return dcmp(Cross(a1 - p, a2 - p)) == 0 && dcmp(Dot(a1 - p, a2 - p)) < 0;
}int ConvexHull(Point* P, int cnt, Point* res) {sort(P, P + cnt);cnt = unique(P, P + cnt) - P;int m = 0;for (int i = 0; i < cnt; i++) {while (m > 1 && Cross(res[m - 1] - res[m - 2], P[i] - res[m - 2]) <= 0)m--;res[m++] = P[i];}int k = m;for (int i = cnt - 2; i >= 0; i--) {while (m > k && Cross(res[m - 1] - res[m - 2], P[i] - res[m - 2]) <= 0)m--;res[m++] = P[i];}if (cnt > 1)m--;return m;
}bool isPointInPolygon(Point p, Point* res, int cnt) {int wn = 0;for (int i = 0; i < cnt; i++) {if (res[i] == p || res[(i + 1) % cnt] == p || OnSegment(p, res[i], res[(i + 1) % cnt]))return 1;int k = dcmp(Cross(res[(i + 1) % cnt] - res[i], p - res[i]));int d1 = dcmp(res[i].y - p.y);int d2 = dcmp(res[(i + 1) % cnt].y - p.y);if (k > 0 && d1 <= 0 && d2 > 0)wn++;if (k < 0 && d2 <= 0 && d1 > 0)wn--;}if (wn)return 1;return 0;
}int main() {while (scanf("%d%d", &n, &m) == 2 && n + m) {int flag = 1;for (int i = 0; i < n; i++)scanf("%lf%lf", &Red[i].x, &Red[i].y);for (int i = 0; i < m; i++)scanf("%lf%lf", &Blue[i].x, &Blue[i].y);int cnt1 = ConvexHull(Red, n, resr);int cnt2 = ConvexHull(Blue, m, resb);for (int i = 0; i < cnt1; i++)if (isPointInPolygon(resr[i], resb, cnt2)) {flag = 0;break;}if (flag) {for (int i = 0; i < cnt2; i++)if (isPointInPolygon(resb[i], resr, cnt1)) {flag = 0;break;}if (flag) {for (int i = 0; i < cnt1; i++) {for (int j = 0; j < cnt2; j++) {if (SegmentProperIntersection(resr[i], resr[(i + 1) % cnt1], resb[j], resb[(j + 1) % cnt2])) {flag = 0;break;}}if (!flag)break;}}}       if (flag)printf("Yes\n");elseprintf("No\n");}return 0;
}

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