POJ3525 Most Distant Point from the Sea

题意：求凸多边形内部离边界最远的点到边界的距离

解法:套用kuangbin的模板，将凸边形一步步向里移动mid距离，判断是否存在核，更改二分条件，直到找到解

#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <set>#include <map>#include <string>#include <math.h>#include <stdlib.h>#include <time.h>using namespace std;const double eps = 1e-8;const double PI = acos(-1.0);int sgn(double x){    if(fabs(x) < eps) return 0;    if(x < 0) return -1;    else return 1;}struct Point{    double x,y;    Point(){}    Point(double _x,double _y)    {        x = _x; y = _y;    }    Point operator -(const Point &b)const    {        return Point(x - b.x, y - b.y);    }    double operator ^(const Point &b)const    {        return x*b.y - y*b.x;    }    double operator *(const Point &b)const    {        return x*b.x + y*b.y;    }};struct Line{    Point s,e;    double k;    Line(){}    Line(Point _s,Point _e)    {        s = _s; e = _e;        k = atan2(e.y - s.y,e.x - s.x);    }    Point operator &(const Line &b)const    {        Point res = s;        double t = ((s - b.s)^(b.s - b.e))/((s - e)^(b.s - b.e));        res.x += (e.x - s.x)*t;        res.y += (e.y - s.y)*t;        return res;    }};//半平面交，直线的左边代表有效区域bool HPIcmp(Line a,Line b){    if(fabs(a.k - b.k) > eps)return a.k < b.k;    return ((a.s - b.s)^(b.e - b.s)) < 0;}Line Q[1010];bool HPI(Line line[], int n){    int tot = n;    sort(line,line+n,HPIcmp);    tot = 1;    for(int i = 1;i < n;i++)        if(fabs(line[i].k - line[i-1].k) > eps)            line[tot++] = line[i];    int head = 0, tail = 1;    Q[0] = line[0];    Q[1] = line[1];    for(int i = 2; i < tot; i++)    {        if(fabs((Q[tail].e-Q[tail].s)^(Q[tail-1].e-Q[tail-1].s)) < eps || fabs((Q[head].e-Q[head].s)^(Q[head+1].e-Q[head+1].s)) < eps)            return false ;        while(head < tail && (((Q[tail]&Q[tail-1]) - line[i].s)^(line[i].e-line[i].s)) > eps)            tail--;        while(head < tail && (((Q[head]&Q[head+1]) - line[i].s)^(line[i].e-line[i].s)) > eps)            head++;        Q[++tail] = line[i];    }    while(head < tail && (((Q[tail]&Q[tail-1]) - Q[head].s)^(Q[head].e-Q[head].s)) > eps)        tail--;    while(head < tail && (((Q[head]&Q[head-1]) - Q[tail].s)^(Q[tail].e-Q[tail].e)) > eps)        head++;    if(tail <= head + 1)return false ;    return true;}Point p[1010];Line line[1010];//*两点间距离double dist(Point a,Point b){    return sqrt((a-b)*(a-b));}void change(Point a,Point b,Point &c,Point &d,double p)//将线段ab往左移动距离p{    double len = dist(a,b);    double dx = (a.y - b.y)*p/len;    double dy = (b.x - a.x)*p/len;    c.x = a.x + dx; c.y = a.y + dy;    d.x = b.x + dx; d.y = b.y + dy;}int main(){    int n;    double r;    while(scanf("%d",&n)&&n)    {        for(int i = 0;i < n;i++)            scanf("%lf%lf",&p[i].x,&p[i].y);        for(int i = 0;i < n;i++)        {            line[i] = Line(p[i],p[(i+1)%n]);        }        double l=0.0,r=20000.0,mid;        while(l+eps<r){            mid=(l+r)/2;            for(int i = 0;i<n;i++)            {                Point t1,t2;              //  printf("qian::::%lf %lf---%lf %lf\n",p[i].x,p[i].y,p[(i+1)%n].x,p[(i+1)%n].y);                change(p[i],p[(i+1)%n],t1,t2,mid);              //  printf("hou::::%lf %lf---%lf %lf\n",t1.x,t1.y,t2.x,t2.y);                line[i] = Line(t1,t2);            }            if(HPI(line,n)) l=mid;            else r=mid;           // printf("l:%lf   %lf\n",l,r);        }        printf("%.6f\n",l);    }    return 0;}