POJ 3525 Most distant point to the sea (half plane and two points) _ two points

Most distant point from the sea Time Limit: 5000MS Memory Limit: 65536K Total submissions: 5136 accepted: 2323 Special Judge Description The main land of the Japan called Honshu is a island surrounded by the sea. In such an island, it's natural to ask a question: "Where's the most distant point from the sea?" The answer to this question for Honshu is found in 1996. The most distant point are located in former Usuda Town, Nagano Prefecture, whose distance to the ' sea ' is 114.86 km. In this problem, your are asked to write a program which, given a map of a island, finds the most EA in the island, and reports its distance to the sea. In order to simplify the problem, we have consider maps representable by convex polygons. Input The input consists of multiple datasets. Each dataset represents a map of the island, which is a convex polygon. The format of a dataset is as follows. N X1 Y1 ⋮ Xn Yn Every input item in a dataset is a non-negative integer. Two input items in a line are separated by a space. N in the ' The ' of the # of vertices of the polygon, satisfying 3≤n≤100. Subsequent n lines are the X-and y-coordinates of the n vertices. Line segments (xi, Yi)-(xi+1, yi+1) (1≤i≤n−1) and the line segment (xn, yn) – (x1, y1) Form the border of the polygon In counterclockwise order. This is the inside of the "polygon in the" the left of their directions of the segments. All coordinate values are between 0 and 10000, inclusive. You can assume this polygon is simple, which is, its border never crosses or touches itself. As stated above, the given polygon is always a convex one. The last dataset was followed by a line containing a single zero. Output For each dataset in the input, one line containing the distance of the the most distant point from the sea should to be output. An output line should not contain extra characters such as spaces. The answer should not have a error greater than 0.00001 (10−5). You could output any number of digits to the decimal point, provided this above accuracy condition is satisfied. Sample Input 4 0 0 10000 0 10000 10000 0 10000 3 0 0 10000 0 7000 1000 6 0/ 100 20< c11/>250 0 3 0 0 10000 10000 5000 5001 0 Sample Output 5000.000000 494.233641 34.542948 0.353553 Source Japan 2007

[Submit] [Go back] [Status] [discuss]

The main idea is to find the farthest distance from the point of convex package to the convex boundary, that is, the maximum incircle of convex polygon.

Solving: Two points + half plane intersection.

Each time two minutes a length, and then let the convex polygon of each edge inward contraction, to the boundary after the contraction to do half plane intersection, if there is no public area or public area for a point, then the length of a large or just, continue to two points.

Mainly internal shrinkage is more difficult to do, I am the adjacent two points of the vector storage, and then in the direction of the normal vector translation to get a new vector, and then a half plane intersection.

#include <iostream> #include <cstdio> #include <algorithm> #include <cstring> #include <
Cmath> #define N #define EPS 1e-7 #define INF 1e9 using namespace std;
	struct vector {double x,y;
	Vector (double x=0,double y=0) {x=x,y=y;
}}a[n],p[n],tmp[n],b[n]; struct data {vector a,b;}
Line[n];
int m,n;
Double Pi=acos (-1.0);
typedef vector Point; 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 t) {return vector (a.x*t,a.y*t);} vector operator/(vector a,double t) {return vector (a.x
/T,A.Y/T);}
	void Init () {m=0;
	P[m++]=point (Inf,inf);
	P[m++]=point (-inf,inf);
	P[m++]=point (-inf,-inf);
P[m++]=point (Inf,-inf);
	int dcmp (double x) {if (Fabs (x) <eps) return 0;
Return x<0?-1:1; Double Cross (vector a,vector b) {return a.x*b.y-a.y*b.x} point GLT (Point a,point a1,point b,point b1) {vector v=a1 -A;
	Vector w=b1-b; Vector u=A-b;
	Double T=cross (w,u)/cross (V,W);
return a+v*t;
	} void Cut (point a,point b) {int cnt=0;
	memset (tmp,0,sizeof (TMP));
		for (int i=0;i<m;i++) {double C=cross (b-a,p[i]-a);
		Double D=cross (b-a,p[(i+1)%m]-a);
		if (dcmp (c) <=0) tmp[cnt++]=p[i];
	if (dcmp (c) *dcmp (d) <0) TMP[CNT++]=GLT (a,b,p[i],p[(i+1)%m]);
	} m=cnt;
for (int i=0;i<cnt;i++) p[i]=tmp[i]; Vector Rotate (vector a,double rad) {return vector (A.x*cos (RAD)-a.y*sin (RAD), A.x*sin (RAD) +a.y*cos (RAD)); Double Len (
		Vector v) {return sqrt (V.X*V.X+V.Y*V.Y);} bool PD (double x) {for (int i=2;i<=n+1;i++) {vector v=b[i]-b[i-1];
		Vector u=rotate (V,PI/2)/len (v);	
		U=u*x;
		LINE[I-1].A=V+U+B[I-1];
		Line[i-1].b=line[i-1].a-v;
	int t=0;
	Init ();
		for (int i=1;i<=n;i++) {cut (line[i].a,line[i].b);
	if (!m) return true;
	} if (m==1) return true;
return false;
	int main () {freopen ("a.in", "R", stdin);
		while (true) {scanf ("%d", &n);
		if (!n) break; for (int i=1;i<=n;i++) scanf ("%lf%lf", &AMP;B[I].X,&AMP;B[I].Y);
		B[N+1]=B[1]; Double l=0; Double r=10000;
		Double ans=0;
			while (r-l>eps) {double mid= (l+r)/2;
			if (PD (mid)) Ans=mid,r=mid-eps;
		else l=mid+eps;
	printf ("%.6lf\n", ans); }
}