POJ 3304 Segments(计算几何)
2015-03-19 11:33
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题意:给定一些线段,问能否找出一条直线,穿过所有线段
思路:如果存在一条直线,那么必然有一条直线是过已有的两点,那么就枚举两点,然后去判断是否跟所有线段有交点即可
代码:
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
struct Point {
double x, y;
Point() {}
Point(double x, double y) {
this->x = x;
this->y = y;
}
void read() {
scanf("%lf%lf", &x, &y);
}
};
typedef Point Vector;
Vector operator + (Vector A, Vector B) {
return Vector(A.x + B.x, A.y + B.y);
}
Vector operator - (Vector A, Vector B) {
return Vector(A.x - B.x, A.y - B.y);
}
Vector operator * (Vector A, double p) {
return Vector(A.x * p, A.y * p);
}
Vector operator / (Vector A, double p) {
return Vector(A.x / p, A.y / p);
}
bool operator < (const Point& a, const Point& b) {
return a.x < b.x || (a.x == b.x && a.y < b.y);
}
const double eps = 1e-8;
int dcmp(double x) {
if (fabs(x) < eps) return 0;
else return x < 0 ? -1 : 1;
}
bool operator == (const Point& a, const Point& b) {
return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0;
}
double Dot(Vector A, Vector B) {return A.x * B.x + A.y * B.y;} //点积
double Length(Vector A) {return sqrt(Dot(A, A));} //向量的模
double Angle(Vector A, Vector B) {return acos(Dot(A, B) / Length(A) / Length(B));} //向量夹角
double Cross(Vector A, Vector B) {return A.x * B.y - A.y * B.x;} //叉积
double Area2(Point A, Point B, Point C) {return Cross(B - A, C - A);} //有向面积
//向量旋转
Vector Rotate(Vector A, double rad) {
return Vector(A.x * cos(rad) - A.y * sin(rad), A.x * sin(rad) + A.y * cos(rad));
}
//判断3点共线
bool LineCoincide(Point p1, Point p2, Point p3) {
return dcmp(Cross(p2 - p1, p3 - p1)) == 0;
}
//判断向量平行
bool LineParallel(Vector v, Vector w) {
return Cross(v, w) == 0;
}
//判断向量垂直
bool LineVertical(Vector v, Vector w) {
return Dot(v, w) == 0;
}
//计算两直线交点,平行,重合要先判断
Point GetLineIntersection(Point P, Vector v, Point Q, Vector w) {
Vector u = P - Q;
double t = Cross(w, u) / Cross(v, w);
return P + v * t;
}
//点到直线距离
double DistanceToLine(Point P, Point A, Point B) {
Vector v1 = B - A, v2 = P - A;
return fabs(Cross(v1, v2)) / Length(v1);
}
//点到线段距离
double DistanceToSegment(Point P, Point A, Point B) {
if (A == B) return Length(P - A);
Vector v1 = B - A, v2 = P - A, v3 = P - B;
if (dcmp(Dot(v1, v2)) < 0) return Length(v2);
else if (dcmp(Dot(v1, v3)) > 0) return Length(v3);
else return fabs(Cross(v1, v2)) / Length(v1);
}
//点在直线上的投影点
Point GetLineProjection(Point P, Point A, Point B) {
Vector v = B - A;
return A + v * (Dot(v, P - A) / Dot(v, v));
}
//线段相交判定(规范相交)
bool SegmentProperIntersection(Point a1, Point a2, Point b1, Point b2) {
double c1 = Cross(a2 - a1, b1 - a1), c2 = Cross(a2 - a1, b2 - a1),
c3 = Cross(b2 - b1, a1 - b1), c4 = Cross(b2 - b1, a2 - b1);
//dcmp(c1) * dcmp(c2) == 0 || dcmp(c3) * dcmp(c4) == 0为不规范相交
return dcmp(c1) * dcmp(c2) <= 0;// && dcmp(c3) * dcmp(c4) <= 0;
}
//判断点在线段上, 不包含端点
bool OnSegment(Point p, Point a1, Point a2) {
return dcmp(Cross(a1 - p, a2 - p)) == 0 && dcmp(Dot(a1 - p, a2 - p)) < 0;
}
//n边形的面积
double PolygonArea(Point *p, int n) {
double area = 0;
for (int i = 1; i < n - 1; i++)
area += Cross(p[i] - p[0], p[i + 1] - p[0]);
return area / 2;
}
const int N = 105;
int t, n;
struct Line {
Point a, b;
void read() {
a.read();
b.read();
}
} line
;
bool judge(Point a, Point b) {
if (dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0) return false;
for (int i = 0; i < n; i++)
if (!SegmentProperIntersection(a, b, line[i].a, line[i].b)) return false;
return true;
}
bool gao() {
for (int i = 0; i < n; i++) {
if (judge(line[i].a, line[i].b)) return true;
for (int j = 0; j < i; j++) {
if (judge(line[i].a, line[j].a)) return true;
if (judge(line[i].a, line[j].b)) return true;
if (judge(line[i].b, line[j].a)) return true;
if (judge(line[i].b, line[j].b)) return true;
}
}
return false;
}
int main() {
scanf("%d", &t);
while (t--) {
scanf("%d", &n);
for (int i = 0; i < n; i++)
line[i].read();
if (gao()) printf("Yes!\n");
else printf("No!\n");
}
return 0;
}
思路:如果存在一条直线,那么必然有一条直线是过已有的两点,那么就枚举两点,然后去判断是否跟所有线段有交点即可
代码:
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
struct Point {
double x, y;
Point() {}
Point(double x, double y) {
this->x = x;
this->y = y;
}
void read() {
scanf("%lf%lf", &x, &y);
}
};
typedef Point Vector;
Vector operator + (Vector A, Vector B) {
return Vector(A.x + B.x, A.y + B.y);
}
Vector operator - (Vector A, Vector B) {
return Vector(A.x - B.x, A.y - B.y);
}
Vector operator * (Vector A, double p) {
return Vector(A.x * p, A.y * p);
}
Vector operator / (Vector A, double p) {
return Vector(A.x / p, A.y / p);
}
bool operator < (const Point& a, const Point& b) {
return a.x < b.x || (a.x == b.x && a.y < b.y);
}
const double eps = 1e-8;
int dcmp(double x) {
if (fabs(x) < eps) return 0;
else return x < 0 ? -1 : 1;
}
bool operator == (const Point& a, const Point& b) {
return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0;
}
double Dot(Vector A, Vector B) {return A.x * B.x + A.y * B.y;} //点积
double Length(Vector A) {return sqrt(Dot(A, A));} //向量的模
double Angle(Vector A, Vector B) {return acos(Dot(A, B) / Length(A) / Length(B));} //向量夹角
double Cross(Vector A, Vector B) {return A.x * B.y - A.y * B.x;} //叉积
double Area2(Point A, Point B, Point C) {return Cross(B - A, C - A);} //有向面积
//向量旋转
Vector Rotate(Vector A, double rad) {
return Vector(A.x * cos(rad) - A.y * sin(rad), A.x * sin(rad) + A.y * cos(rad));
}
//判断3点共线
bool LineCoincide(Point p1, Point p2, Point p3) {
return dcmp(Cross(p2 - p1, p3 - p1)) == 0;
}
//判断向量平行
bool LineParallel(Vector v, Vector w) {
return Cross(v, w) == 0;
}
//判断向量垂直
bool LineVertical(Vector v, Vector w) {
return Dot(v, w) == 0;
}
//计算两直线交点,平行,重合要先判断
Point GetLineIntersection(Point P, Vector v, Point Q, Vector w) {
Vector u = P - Q;
double t = Cross(w, u) / Cross(v, w);
return P + v * t;
}
//点到直线距离
double DistanceToLine(Point P, Point A, Point B) {
Vector v1 = B - A, v2 = P - A;
return fabs(Cross(v1, v2)) / Length(v1);
}
//点到线段距离
double DistanceToSegment(Point P, Point A, Point B) {
if (A == B) return Length(P - A);
Vector v1 = B - A, v2 = P - A, v3 = P - B;
if (dcmp(Dot(v1, v2)) < 0) return Length(v2);
else if (dcmp(Dot(v1, v3)) > 0) return Length(v3);
else return fabs(Cross(v1, v2)) / Length(v1);
}
//点在直线上的投影点
Point GetLineProjection(Point P, Point A, Point B) {
Vector v = B - A;
return A + v * (Dot(v, P - A) / Dot(v, v));
}
//线段相交判定(规范相交)
bool SegmentProperIntersection(Point a1, Point a2, Point b1, Point b2) {
double c1 = Cross(a2 - a1, b1 - a1), c2 = Cross(a2 - a1, b2 - a1),
c3 = Cross(b2 - b1, a1 - b1), c4 = Cross(b2 - b1, a2 - b1);
//dcmp(c1) * dcmp(c2) == 0 || dcmp(c3) * dcmp(c4) == 0为不规范相交
return dcmp(c1) * dcmp(c2) <= 0;// && dcmp(c3) * dcmp(c4) <= 0;
}
//判断点在线段上, 不包含端点
bool OnSegment(Point p, Point a1, Point a2) {
return dcmp(Cross(a1 - p, a2 - p)) == 0 && dcmp(Dot(a1 - p, a2 - p)) < 0;
}
//n边形的面积
double PolygonArea(Point *p, int n) {
double area = 0;
for (int i = 1; i < n - 1; i++)
area += Cross(p[i] - p[0], p[i + 1] - p[0]);
return area / 2;
}
const int N = 105;
int t, n;
struct Line {
Point a, b;
void read() {
a.read();
b.read();
}
} line
;
bool judge(Point a, Point b) {
if (dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0) return false;
for (int i = 0; i < n; i++)
if (!SegmentProperIntersection(a, b, line[i].a, line[i].b)) return false;
return true;
}
bool gao() {
for (int i = 0; i < n; i++) {
if (judge(line[i].a, line[i].b)) return true;
for (int j = 0; j < i; j++) {
if (judge(line[i].a, line[j].a)) return true;
if (judge(line[i].a, line[j].b)) return true;
if (judge(line[i].b, line[j].a)) return true;
if (judge(line[i].b, line[j].b)) return true;
}
}
return false;
}
int main() {
scanf("%d", &t);
while (t--) {
scanf("%d", &n);
for (int i = 0; i < n; i++)
line[i].read();
if (gao()) printf("Yes!\n");
else printf("No!\n");
}
return 0;
}
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