A Round Peg in a Ground Hole
Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 5456 Accepted: 1735 Description
The DIY Furniture company specializes in assemble-it-yourself furniture kits. Typically, the pieces of wood are attached to one another using a wooden peg that fits into pre-cut holes in each piece to be attached. The pegs have a circular cross-section and so are intended to fit inside a round hole.
A recent factory run of computer desks were flawed when an automatic grinding machine was mis-programmed. The result is an irregularly shaped hole in one piece that, instead of the expected circular shape, is actually an irregular polygon. You need to figure out whether the desks need to be scrapped or if they can be salvaged by filling a part of the hole with a mixture of wood shavings and glue.
There are two concerns. First, if the hole contains any protrusions (i.e., if there exist any two interior points in the hole that, if connected by a line segment, that segment would cross one or more edges of the hole), then the filled-in-hole would not be structurally sound enough to support the peg under normal stress as the furniture is used. Second, assuming the hole is appropriately shaped, it must be big enough to allow insertion of the peg. Since the hole in this piece of wood must match up with a corresponding hole in other pieces, the precise location where the peg must fit is known.
Write a program to accept descriptions of pegs and polygonal holes and determine if the hole is ill-formed and, if not, whether the peg will fit at the desired location. Each hole is described as a polygon with vertices (x1, y1), (x2, y2), . . . , (xn, yn). The edges of the polygon are (xi, yi) to (xi+1, yi+1) for i = 1 . . . n − 1 and (xn, yn) to (x1, y1).Input
Input consists of a series of piece descriptions. Each piece description consists of the following data:
Line 1 < nVertices > < pegRadius > < pegX > < pegY >
number of vertices in polygon, n (integer)
radius of peg (real)
X and Y position of peg (real)
n Lines < vertexX > < vertexY >
On a line for each vertex, listed in order, the X and Y position of vertex The end of input is indicated by a number of polygon vertices less than 3.Output
For each piece description, print a single line containing the string:
HOLE IS ILL-FORMED if the hole contains protrusions
PEG WILL FIT if the hole contains no protrusions and the peg fits in the hole at the indicated position
PEG WILL NOT FIT if the hole contains no protrusions but the peg will not fit in the hole at the indicated positionSample Input
5 1.5 1.5 2.0 1.0 1.0 2.0 2.0 1.75 2.0 1.0 3.0 0.0 2.0 5 1.5 1.5 2.0 1.0 1.0 2.0 2.0 1.75 2.5 1.0 3.0 0.0 2.0 1Sample Output
HOLE IS ILL-FORMED PEG WILL NOT FITSource
这道题其实并不难，难得是独栋题意，这个题比较难懂，可能是英语比较差吧，读了好久。。。
题目大意： 给定n个点，但是不确定是顺时针还是逆时针的，判断它是否是凸多边形，如果是凸度变形的话，题目给定了一个圆心和半径，让求圆是否在凸多边形内。
关键步骤：
1. 判断是否为凸多边形
2. 如果是凸多边形，判断圆心在不在多边形内(边上也算)
3. 如果圆心在，判断整个圆在不在，这时候就比较圆心到各个边的距离与半径的关系就行了
下面是用叉积来判断是否是凸多边形，用向量的点乘积来判断圆是否在多边形内，代码如下：
/************************************************************************* > File Name: poj_1584_back.cpp > Author: Howe_Young > Mail: 1013410795@qq.com > Created Time: 2015年04月06日 星期一 19时03分03秒 ************************************************************************/ #include <cstdio> #include <iostream> #include <cstring> #include <cmath> #include <cstdlib> #include <algorithm> #include <cstdio> #define EPS 1e-8 using namespace std; struct point{ double x, y; }; const int N = 1000; point p[N]; double Min(double a, double b) { return a < b ? a : b; } double Max(double a, double b) { return a > b ? a : b; } int sgn(double d) { if (fabs(d) < EPS) return 0; return d > 0 ? 1 : -1; } double x_mutli(point p1, point p2) { return (p1.x * p2.y - p2.x * p1.y); } double dot_multi(point p1, point p2) { return (p1.x * p2.x + p1.y * p2.y); } //方向 double get_direction(point p1, point p2, point p3) { point v1, v2; v1.x = p3.x - p1.x; v1.y = p3.y - p1.y; v2.x = p2.x - p1.x; v2.y = p2.y - p1.y; return x_mutli(v1, v2); } //得到模长 double get_length_of_mold(point p) { return sqrt(p.x * p.x + p.y * p.y); } bool is_vonvex(int n) { double tmp1 = 0.0, tmp2; for (int i = 0; i < n; i++)//因为凸多边形的相邻边的拐向都相同,要么都顺时针,要么多逆时针 { tmp2 = sgn(get_direction(p[i], p[(i+1)%n], p[(i+2)%n])); if (tmp1 * tmp2 < -EPS) return false; tmp1 = tmp2; } return true; } //叉积判断点是否在多边形内,只适合凸多边形 bool point_in_polygon(point peg, int n) { double tmp1 = 0.0, tmp2; for (int i = 0; i < n; i++) { tmp2 = sgn(get_direction(p[i], p[(i+1)%n], peg)); if (tmp1 * tmp2 < -EPS) return false; tmp1 = tmp2; } return true; } //判断圆是否在多边形内,就是判断点到边的最小离跟半径的关系 bool circle_in_polygon(point peg, double peg_r, int n) { if (peg_r == 0.0) return true; double shadow;//v1向量在v2向量上的投影长度 a点乘b然后除以b的模就是a在b上的投影 point v1, v2; double ans; for (int i = 0; i < n; i++) { v1.x = peg.x - p[i].x; v1.y = peg.y - p[i].y; v2.x = p[(i+1)%n].x - p[i].x; v2.y = p[(i+1)%n].y - p[i].y; shadow = dot_multi(v1, v2) / (v2.x * v2.x + v2.y * v2.y) * 1.0;//这里是求投影占v2向量模的长度 的比例,如果大于1,或者小于0, 垂足肯定在外面了 if (shadow >= 0.0 && shadow <= 1.0)//利用面积来求高,也就是距离,叉乘的绝对值是三角形面积的两倍 ans = fabs(x_mutli(v1, v2)) / get_length_of_mold(v2); else { //如果垂足在外面,找最近的一个端点 v2.x = peg.x - p[(i+1)%n].x; v2.y = peg.y - p[(i+1)%n].y; ans = sgn(get_length_of_mold(v1) - get_length_of_mold(v2)) == -1 ? get_length_of_mold(v1) : get_length_of_mold(v2); } if (ans - peg_r < -EPS)//如果相交,返回false return false; } return true; } int main() { //freopen("in.txt", "r", stdin); //freopen("out.txt", "w", stdout); int n; point peg; double peg_r; while (~scanf("%d", &n) && n >= 3) { scanf("%lf %lf %lf", &peg_r, &peg.x, &peg.y); for (int i = 0; i < n; i++) scanf("%lf %lf", &p[i].x, &p[i].y); if (!is_vonvex(n))//判断是否是凸多边形 { puts("HOLE IS ILL-FORMED"); continue; } if (point_in_polygon(peg , n) && circle_in_polygon(peg, peg_r, n)) { puts("PEG WILL FIT"); } else puts("PEG WILL NOT FIT"); } return 0; }
用射线的方法来判断点是否在多边形内
bool on_line(point p1, point p2, point p3) { if (p3.x >= Min(p1.x, p2.x) && p3.x <= Max(p1.x, p2.x) && p3.y <= Min(p1.y, p2.y) && p3.y <= Max(p1.y, p2.y)) return sgn(get_direction(p1, p2, p3)) == 0; return false; } //射线的方法判别点在多边形内 bool point_in_polygon_ray(point peg, int n, double peg_r) { int counter = 0; double xinter; point p1, p2; p1 = p[0]; for (int i = 1; i <= n; i++) { p2 = p[i % n]; if (on_line(p1, p2, peg)) if (sgn(peg_r) == 0) return false; else return true; if (peg.y > Min(p1.y, p2.y)) { if (peg.y <= Max(p1.y, p2.y)) { if (peg.x <= Max(p1.x, p2.x)) { if (p1.y != p2.y) { xinter = (peg.y - p2.y) * (p1.x - p2.x) / (p1.y - p2.y) + p2.x; if (p1.x == p2.x || peg.x <= xinter) counter++; } } } } p1 = p2; } if (counter % 2 == 0) return false; return true; }
角度和判定是否在多边形内
//判断点在直线上 p3在p1p2上 bool point_on_line(point p1, point p2, point p3) { if (p3.x >= Min(p1.x, p2.x) && p3.x <= Max(p1.x, p2.x) && p3.y >= Min(p1.y, p2.y) && p3.y <= Max(p1.y, p2.y)) return sgn(get_direction(p1, p2, p3)) == 0; return false; } //判断点是否在多边形内，角度和算法 bool point_is_inside_angle(point peg, int n) { double sum = 0.0; point v1, v2, v3; for (int i = 0; i < n; i++) { if (peg == p[i]) return true; if (p[i] == p[(i + 1) % n]) continue; if (point_on_line(p[i], p[(i+1)%n], peg)) return true; v1.x = peg.x - p[i].x; v1.y = peg.y - p[i].y; v2.x = peg.x - p[(i+1)%n].x; v2.y = peg.y - p[(i+1)%n].y; v3.x = p[i].x - p[(i+1)%n].x; v3.y = p[i].y - p[(i+1)%n].y; double a = get_length_of_mold(v1); double b = get_length_of_mold(v2); double c = get_length_of_mold(v3); sum += sgn(x_mutli(v1, v2)) * acos((a * a + b * b - c * c) / (2.0 * a * b)); } sum = fabs(sum); if (sgn(sum - 2.0 * PI) == 0) return true; return false; }
改进弧长法(这个方法还未理解，如果哪位大神路过这，麻烦留一下言)
double x_multi_2(point p1,point p2,point p3)
{
return (p2.x-p1.x)*(p3.y-p1.y)-(p3.x-p1.x)*(p2.y-p1.y);
}
int get_tmp(point p0)
{
return p0.x>=0?(p0.y>=0?0:3):(p0.y>=0?1:2);
}
//改进弧长法
bool point_is_inside_arc(point peg, int n)
{
int tmp1,tmp2,sum=0,i;
point p0,p1;
p0.x=peg.x,p0.y=peg.y;
p1.x=p[0].x-p0.x,p1.y=p[0].y-p0.y;
tmp1=get_tmp(p1);
for(i=0;i<n;i++)
{
if(p[i]==p0)
break;
int t0=sgn(x_multi_2(p[i],p[(i+1)%n], p0));
int t1=sgn((p[i].x-p0.x)*(p[(i+1)%n].x-p0.x));
int t2=sgn((p[i].y-p0.y)*(p[(i+1)%n].y-p0.y));
if(!t0&&t1<=0&&t2<=0) //被测点在多边形边上
break;
p1.x=p[(i+1)%n].x-p0.x,p1.y=p[(i+1)%n].y-p0.y;
tmp2=get_tmp(p1); //计算象限
switch((tmp2-tmp1+4)%4)
{
case 1:{ sum++; break; }
case 2:
{
if(t0>0) sum+=2;
else sum-=2;
break;
}
case 3: { sum--; break; }
}
tmp1=tmp2;
}
if(i<n||sum) //被测点在多边形边上或者在多边形内部
return true;
return false;
}