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POJ 3525 Most Distant Point from the Sea [半平面交 二分]
| Time Limit: 5000MS | Memory Limit: 65536K | |||
| Total Submissions: 5153 | Accepted: 2326 | Special Judge | ||
Description
The main land of Japan called Honshu is an island surrounded by the sea. In such an island, it is natural to ask a question: “Where is the most distant point from the sea?” The answer to this question for Honshu was found in 1996. The most distant point is located in former Usuda Town, Nagano Prefecture, whose distance from the sea is 114.86 km.
In this problem, you are asked to write a program which, given a map of an island, finds the most distant point from the sea in the island, and reports its distance from the sea. In order to simplify the problem, we only consider maps representable by convex polygons.
Input
The input consists of multiple datasets. Each dataset represents a map of an 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 first line is the number 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. That is, these line segments see the inside of the polygon in the left of their directions. All coordinate values are between 0 and 10000, inclusive.
You can assume that the polygon is simple, that is, its border never crosses or touches itself. As stated above, the given polygon is always a convex one.
The last dataset is followed by a line containing a single zero.
Output
For each dataset in the input, one line containing the distance of the most distant point from the sea should be output. An output line should not contain extra characters such as spaces. The answer should not have an error greater than 0.00001 (10−5). You may output any number of digits after the decimal point, provided that the above accuracy condition is satisfied.
Sample Input
40 010000 010000 100000 1000030 010000 07000 100060 40100 20250 40250 70100 900 7030 010000 100005000 50010
Sample Output
5000.000000494.23364134.5429480.353553
Source
题意：
给出一个形状为凸包的岛屿，求岛上离海(即凸包外)最远的点离海的距离有多远二分多远，然后把凸包缩小这么远，看看此时半平面交有没有交集
#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <vector>using namespace std;typedef long long ll;const int N=105;const double INF=1e4+5;const double eps=1e-10;inline int read(){ char c=getchar();int x=0,f=1; while(c<‘0‘||c>‘9‘){if(c==‘-‘)f=-1; c=getchar();} while(c>=‘0‘&&c<=‘9‘){x=x*10+c-‘0‘; c=getchar();} return x*f;}inline int sgn(double x){ if(abs(x)<eps) return 0; else return x<0?-1:1;}struct Vector{ double x,y; Vector(double a=0,double b=0):x(a),y(b){} bool operator <(const Vector &a)const{ return sgn(x-a.x)<0||(sgn(x-a.x)==0&&sgn(y-a.y)<0); } void print(){printf("%lf %lf\n",x,y);}};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 b){return Vector(a.x*b,a.y*b);}Vector operator /(Vector a,double b){return Vector(a.x/b,a.y/b);}bool operator ==(Vector a,Vector b){return sgn(a.x-b.x)==0&&sgn(a.y-b.y)==0;}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 Len(Vector a){return sqrt(Dot(a,a));}Vector Normal(Vector a){ return Vector(-a.y,a.x);//counterClockwise}struct Line{ Point s,t; Line(){} Line(Point a,Point b):s(a),t(b){}};bool isLSI(Line l1,Line l2){ Vector v=l1.t-l1.s,u=l2.s-l1.s,w=l2.t-l1.s; return sgn(Cross(v,u))!=sgn(Cross(v,w));}Point LI(Line a,Line b){ Vector v=a.s-b.s,v1=a.t-a.s,v2=b.t-b.s; double t=Cross(v2,v)/Cross(v1,v2); return a.s+v1*t;}void iniPolygon(Point p[],int &n,double inf){ n=0; p[++n]=Point(inf,inf); p[++n]=Point(inf,-inf); p[++n]=Point(-inf,-inf); p[++n]=Point(-inf,inf);}Point t[N];int tn;void CutPolygon(Point p[],int &n,Point a,Point b){//get the left of a->b tn=0; Point c,d; for(int i=1;i<=n;i++){ c=p[i],d=p[i%n+1]; if(sgn(Cross(b-a,c-a))>=0) t[++tn]=c; if(isLSI(Line(a,b),Line(c,d))) t[++tn]=LI(Line(a,b),Line(c,d)); } n=tn;for(int i=1;i<=n;i++) p[i]=t[i];}int n,m;Point p[N],q[N];Line L[N];void ChangePolygon(Point p[],int n,double x){ p[n+1]=p[1]; for(int i=1;i<=n;i++){ Vector v=Normal(p[i+1]-p[i])*x/Len(p[i+1]-p[i]); L[i]=Line(p[i]+v,p[i+1]+v); }}void solve(){ double l=0,r=10000,e=1e-6; while(r-l>e){ double mid=(l+r)/2;//printf("hi %lf %lf %lf\n",l,r,mid); ChangePolygon(p,n,mid); iniPolygon(q,m,INF); for(int i=1;i<=n;i++) CutPolygon(q,m,L[i].s,L[i].t); if(m) l=mid; else r=mid; } printf("%lf\n",l);}int main(int argc, const char * argv[]){ while(true){ n=read();if(n==0) break; for(int i=1;i<=n;i++) scanf("%lf%lf",&p[i].x,&p[i].y); solve(); }}
POJ 3525 Most Distant Point from the Sea [半平面交 二分]