POJ3525-Most Distant Point from the Sea（二分+半平面交）

Most Distant Point from the Sea

Time Limit: 5000MS Memory Limit: 65536KTotal Submissions: 3955 Accepted: 1847 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 ⋮ x_n y_n

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 (x_i, y_i)–(x_i+1, y_i+1) (1 ≤ i ≤ n − 1) and the line segment (x_n, y_n)–(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

题意：让你求凸包内的点距离凸包边上最大的距离。

思路：可以二分答案，求半平面交。

#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-9;const int maxn = 100*2+10;struct Point{double x,y;Point(double x=0.0,double y = 0.0):x(x),y(y){}};vector<Point> vP;typedef Point Vector;Point poly[maxn];vector<Vector> vV1,vV2;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 L[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);}bool operator < (const Point &a,const Point &b){return a.x < b.x || (a.x==a.y && a.y < b.y);}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.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 OnLeft(Line L,Point p){return Cross(L.v,p-L.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 HalfplaneIntersection(Line* L,int n,Point* poly){sort(L,L+n);int first,last;Point *p = new Point[n];Line *q = new Line[n];q[first=last=0] = L[0];for(int i = 1; i < n; i++){while(first < last && !OnLeft(L[i],p[last-1])) last--;while(first < last && !OnLeft(L[i],p[first])) first++;q[++last] = L[i];if(fabs(Cross(q[last].v,q[last-1].v))<eps) {last--;if(OnLeft(q[last],L[i].P)) q[last] = L[i];}if(first<last) p[last-1] = GetIntersection(q[last-1],q[last]);}while(first < last && !OnLeft(q[first],p[last-1])) last--;if(last - first <=1) return 0;p[last] = GetIntersection(q[last],q[first]);int m = 0;for(int i = first; i <= last; i++) poly[m++] = p[i];return m;}int n;void init(){vP.clear();vV1.clear();vV2.clear();}void input(){REP(_,0,n){double x,y;scanf("%lf%lf",&x,&y);vP.push_back(Point(x,y));}#define next(i) ((i)+1)%nREP(_,0,n){vV1.push_back(vP[next(_)]-vP[(_)]);vV2.push_back(Normal(vP[next(_)]-vP[(_)]));}}void solve(){double l = 0,r = 40000;while(r-l > eps){double mid = (l+r)/2;REP(_,0,n) {L[_] = Line(vP[_]+vV2[_]*mid ,vV1[_]);}int m = HalfplaneIntersection(L,n,poly);if(m==0){r = mid;}else{l = mid;}}printf("%.7f\n",r);}int main(){while(~scanf("%d",&n) && n){init();input();solve();}return 0;}