一个多边形内部有3枚钉子_多边形的内部和外部
一个多边形内部有3枚钉子_多边形的内部和外部
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Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other) Total Submission(s) : 24 Accepted Submission(s) : 7 Problem Description Statement of the Problem Several drawing applications allow us to draw polygons and almost all of them allow us to fill them with some color. The task of filling a polygon reduces to knowing which points are inside it, so programmers have to colour only those points. You’re expected to write a program which tells us if a given point lies inside a given polygon described by the coordinates of its vertices. You can assume that if a point is in the border of the polygon, then it is in fact inside the polygon.
Input Format
The input file may contain several instances of the problem. Each instance consists of: (i) one line containing integers N, 0 < N < 100 and M, respectively the number of vertices of the polygon and the number of points to be tested. (ii) N lines, each containing a pair of integers describing the coordinates of the polygon’s vertices; (iii) M lines, each containing a pair of integer coordinates of the points which will be tested for “withinness” in the polygon.
You may assume that: the vertices are all distinct; consecutive vertices in the input are adjacent in the polygon; the last vertex is adjacent to the first one; and the resulting polygon is simple, that is, every vertex is incident with exactly two edges and two edges only intersect at their common endpoint. The last instance is followed by a line with a 0 (zero).
Output Format
For the ith instance in the input, you have to write one line in the output with the phrase “Problem i:”, followed by several lines, one for each point tested, in the order they appear in the input. Each of these lines should read “Within” or “Outside”, depending on the outcome of the test. The output of two consecutive instances should be separated by a blank line.
Sample Input
3 1
0 0
0 5
5 0
10 2
3 2
4 4
3 1
1 2
1 3
2 2
0题解 判断点在多边形内部
#include<bits/stdc++.h>
using namespace std;
const double eps = 1e-8;
double pi = acos(-1);
int sgn(double x){
if(fabs(x) < eps)return 0;
if(x < 0)return -1;
return 1;
}
int dcmp(double x,double y){
if(fabs(x - y) < eps)return 0;
if(x < y)return -1;
return 1;
}
struct Point{
double x,y;
Point(){
}
Point(double x,double y):x(x),y(y){
}
Point operator+(Point B){
return Point(x + B.x,y +B.y);}
Point operator-(Point B){
return Point(x - B.x,y - B.y);}
Point operator*(double k){
return Point(x * k,y * k);}
Point operator/(double k){
return Point(x / k,y / k);}
bool operator==(Point B){
return !dcmp(x,B.x) && !dcmp(y,B.y);}
};
typedef Point Vector;
double Dot(Vector A,Vector B){
return A.x * B.x + A.y * B.y;
}
double Cross(Vector A,Vector B){
return A.x * B.y - A.y * B.x;
}
double Distance(Point A,Point B){
return sqrt((A.x - B.x) * (A.x - B.x) + (A.y - B.y) * (A.y - B.y));
}
struct Line{
Point p1,p2;
Line(){
}
Line(Point p1,Point p2):p1(p1),p2(p2){
}
Line(Point p,double angle){
p1 = p;
if(sgn(angle - pi / 2) == 0)p2 = (p1 + Point(0,1));
else p2 = (p1 + Point(1,tan(angle)));
}
Line(double a,double b,double c){
if(sgn(a) == 0)p1 = Point(0,-c / b),p2 = Point(1,-c / b);
else if(sgn(b) == 0)p1 = Point(-c / a,0),p2 = Point(-c / a,1);
else p1 = Point(0,-c / b),p2 = Point(1,(-c - a) / b);
}
};
typedef Line Segment;
int Point_Lint_Relation(Point p,Line v){
int c = sgn(Cross(p - v.p1,v.p2 - v.p1));
if(c == -1)return 1;//点在直线的左边
else if(c == 0)return 0;//点在直线上
else return 2;//点在直线的右边
}
bool Point_On_Line(Point p,Segment v){
return sgn(Cross(p - v.p1,p - v.p2)) == 0 && Dot(v.p1 - p,v.p2 - p) <= 0;
}
bool Point_In_Polygon(Point pt,Point *p,int n){
for(int i = 0;i < n;i ++)
if(p[i] == pt)return true;
for(int i = 0;i < n;i ++){
Line v(p[i],p[(i + 1) % n]);
if(Point_On_Line(pt,v))return true;
}
int num = 0;
for(int i = 0;i < n;i ++){
Point p1 = p[i],p2 = p[(i + 1) % n];
int dir = Cross(p2 - p1,pt - p1);
int u = p1.y - pt.y,v = p2.y - pt.y;
if(dir > 0 && sgn(u) < 0 && sgn(v) >= 0)num ++;
else if(dir < 0 && sgn(u) >= 0 && sgn(v) < 0)num --;
}
return num != 0;
}
const int N = 110;
Point polygons[N];
int main()
{
int n,m;
int x,y;
int T = 0;
while(cin>>n,n){
cin>>m;
if(T != 0)printf("\n");
printf("Problem %d:\n", ++T);
for(int i = 0;i < n;i ++){
cin>>x>>y;
polygons[i].x = x,polygons[i].y = y;
}
for(int i = 0;i < m;i ++){
cin>>x>>y;
Point pt(x,y);
if(Point_In_Polygon(pt,polygons,n))cout<<"Within"<<endl;
else cout<<"Outside"<<endl;
}
}
return 0;
}发布者：全栈程序员栈长，转载请注明出处：https://javaforall.cn/169110.html原文链接：https://javaforall.cn
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