Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 5301 | Accepted: 1662 |
Description
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
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
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 position
Sample 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 1
Sample Output
HOLE IS ILL-FORMED PEG WILL NOT FIT
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <vector>
#include <algorithm>
using namespace std;
#define REP(_,a,b) for(int _ = (a); _ < (b); _++)
#define sz(s) (int)((s).size())
typedef long long ll;
const double eps = 1e-10;
const int maxn = 50;
int n;
struct Point{
double x,y;
Point(double x=0.0,double y = 0.0):x(x),y(y){}
};
Point vP[maxn];
typedef Point Vector;
struct Line {
Point P;
Vector v;
double ang;
Line(){}
Line(Point P,Vector v):P(P),v(v){
ang = atan2(v.y,v.x);
}
bool operator <(const Line&L) const{
return ang < L.ang;
}
};
Line LX[maxn],LY[maxn];
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);
}
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.x-b.x)==0&&dcmp(a.y-b.y)<0;
}
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;}
double Length(Vector A) {return sqrt(Dot(A,A));}
double Angle(Vector A,Vector B) {return acos(Dot(A,B)/Length(A)/Length(B));}
double Cross(Vector A,Vector B) {return A.x*B.y-A.y*B.x;}
Vector Rotate(Vector A,double rad) {return Vector(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad)); }
Vector Normal(Vector A) {
double L = Length(A);
return Vector(-A.y/L,A.x/L);
}
bool OnSegment(Point p,Point a1,Point a2){
return dcmp(Cross(a1-p,a2-p)) == 0 && dcmp(Dot(a1-p,a2-p)) < 0;
}
Point GetIntersection(Line a,Line b){
Vector u = a.P-b.P;
double t = Cross(b.v,u) / Cross(a.v,b.v);
return a.P+a.v*t;
}
int ConvexHull(Point* p,int n,Point *ch){
sort(p,p+n);
int m = 0;
for(int i = 0; i < n; i++) {
while(m > 1 && dcmp(Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])) <= 0) m--;
ch[m++] = p[i];
}
int k = m;
for(int i = n-2; i >= 0; i--){
while(m > k && dcmp(Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])) <= 0) m--;
ch[m++] = p[i];
}
if(n > 1) m--;
return m;
}
double a[maxn];
void input(){
double t;
REP(i,0,n) scanf("%lf",&a[i]);
LX[0] = Line(Point(0,1.0),Vector(0,-1.0));
REP(i,0,n){
scanf("%lf",&t);
LX[i+1] = Line(Point(t,1.0),Vector(a[i]-t,-1.0));
}
LX[n+1] = Line(Point(1.0,1.0),Vector(0,-1.0));
LY[0] = Line(Point(1.0,0),Vector(-1.0,0));
REP(i,0,n) scanf("%lf",&a[i]);
REP(i,0,n){
scanf("%lf",&t);
LY[i+1] = Line(Point(1.0,t),Vector(-1.0,a[i]-t));
}
LY[n+1] = Line(Point(1.0,1.0),Vector(-1.0,0));
}
void solve(){
double ans = 0;
REP(i,1,n+2) {
Point a,b,c,d;
REP(j,1,n+2) {
a = GetIntersection(LX[i-1],LY[j-1]);
//cout<<a.x<<" "<<a.y<<endl;
b = GetIntersection(LX[i],LY[j-1]);
//cout<<b.x<<" "<<a.y<<endl;
c = GetIntersection(LX[i],LY[j]);
d = GetIntersection(LX[i-1],LY[j]);
double area = 0;
area = fabs(Cross(a,b)/2+Cross(b,c)/2+Cross(c,d)/2+Cross(d,a)/2);
ans = max(area,ans);
}
}
printf("%.6f\n",ans);
}
int main(){
while(~scanf("%d",&n) && n){
input();
solve();
}
return 0;
}