uvalive 3890

题意：有一个n个点的凸包多边形，在这个多边形内找一点到多边形边界的距离最远，输出最远距离。
题解：利用半平面交求多边形的内核，二分出距离，多边形的每条边都按这个距离向内推进，如果有交点就继续推进。
http://blog.csdn.net/accry/article/details/6070621这里的半平面交讲的不错。

#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>using namespace std;const double PI = acos(-1);const int N = 105;const double INF = 1e9;const double eps = 1e-7;struct Point {    double x, y;    Point(double x = 0, double y = 0): x(x), y(y) {}}P[N], poly[N], v1[N], v2[N];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;    }}L[N];int n;double Sqr(double x) {    return x * x;}Point operator + (Point A, Point B) {    return Point(A.x + B.x, A.y + B.y);}Point operator - (Point A, Point B) {    return Point(A.x - B.x, A.y - B.y);}Point operator * (Point A, double p) {    return Point(A.x * p, A.y * p);}Point operator / (Point A, double p) {    return Point(A.x / p, A.y / p);}//计算点积的正负  负值夹角为钝角int dcmp(double x) {    if (fabs(x) < 1e-9)        return 0;    return x < 0 ? -1 : 1;}bool operator < (const Point& a, const Point& b) {    return a.x < b.x || (a.x == b.x && a.y < b.y);}bool operator == (const Point& a, const Point& b) {    return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0;}//计算点积double Dot(Point A, Point B) {    return A.x * B.x + A.y * B.y;}//计算叉积，也就是数量积double Cross(Point A, Point B) {    return A.x * B.y - A.y * B.x;}//计算向量长度double Length(Point A) {    return sqrt(Dot(A, A));}Vector Normal(Vector A) {    double L = Length(A);    return Vector(-A.y / L, A.x / L);}//向量A旋转rad弧度，rad负值为顺时针旋转Vector Rotate(Vector A, double rad) {    return Vector(A.x * cos(rad) - A.y * sin(rad), A.x * sin(rad) + A.y * cos(rad));}//角度转化弧度double torad(double deg) {    return deg / 180.0 * PI;}//点p在有向直线L的左边(线上不算)bool OnLeft(Line L, Point P) {    return Cross(L.v, P - L.p) > 0;}//求两直线的交点，前提交点一定存在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;}//求半面交int HalfplaneIntersection(Line* L, int n, Point* poly) {    sort(L, L + n);    int first = 0, rear = 0;    Point *p = new Point[n];    Line *q = new Line[n];    q[first] = L[0];    for (int i = 1; i < n; i++) {        while (first < rear && !OnLeft(L[i], p[rear - 1]))            rear--;        while (first < rear && !OnLeft(L[i], p[first]))            first++;        q[++rear] = L[i];        if (fabs(Cross(q[rear].v, q[rear - 1].v)) < eps) {            rear--;            if (OnLeft(q[rear], L[i].p))                q[rear] = L[i];        }        if (first < rear)            p[rear - 1] = GetIntersection(q[rear - 1], q[rear]);    }    while (first < rear && !OnLeft(q[first], p[rear - 1]))        rear--;     if (rear - first <= 1)        return 0;    p[rear] = GetIntersection(q[rear], q[first]);    int m = 0;    for (int i = first; i <= rear; i++)        poly[m++] = p[i];    return m;}int main() {    while (scanf("%d", &n) == 1 && n) {        for (int i = 0; i < n; i++)            scanf("%lf%lf", &P[i].x, &P[i].y);        for (int i = 0; i < n; i++) {            v1[i] = P[(i + 1) % n] - P[i];            v2[i] = Normal(v1[i]);        }        double left = 0, right = 20000;        while (right - left > eps) {            double mid = (left + right) / 2;                for (int i = 0; i < n; i++)                L[i] = Line(P[i] + v2[i] * mid, v1[i]);            int m = HalfplaneIntersection(L, n, poly);            if (!m)                right = mid;            else                left = mid;        }        printf("%lf\n", left);    }    return 0;}