【POJ3525】Most Distant Point from the Sea-半平面交+二分答案

测试地址：Most Distant Point from the Sea
题目大意：给定一个凸多边形，求凸多边形内某点到各边的最小距离的最大值。
做法：本题需要用到半平面交+二分答案。
首先答案显然具有单调性，所以我们二分答案，转化成判定性问题。
考虑答案d$d$，我们只要将构成原凸多边形的半平面都向内缩d$d$个单位长度，然后判定半平面交存不存在即可。因为不用排序，所以时间复杂度为O(n)$O(n)$，配上外面的二分就多个log$\log$。
以下是本人代码：

#include <bits/stdc++.h>
#define eps 1e-12
using namespace std;
int n,bot,top;
struct Point
{
    double x,y;
}p[110],zero;
struct Line
{
    Point a,b;
}l[110],q[110];

Point operator - (Point a,Point b)
{
    Point s={a.x-b.x,a.y-b.y};
    return s;
}

double dis(Point a,Point b)
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}

double multi(Point a,Point b)
{
    return a.x*b.y-b.x*a.y;
}

Point inter(Line a,Line b)
{
    double a1=multi(b.b-a.a,b.a-a.a),a2=multi(b.a-a.b,b.b-a.b);
    Point s={(a2*a.a.x+a1*a.b.x)/(a2+a1),(a2*a.a.y+a1*a.b.y)/(a2+a1)};
    return s;
}

bool JudgeOut(Point a,Line b)
{
    return multi(a-b.a,b.b-b.a)>0.0;
}

bool solve()
{
    q[1]=l[1],q[2]=l[2],bot=1,top=2;
    for(int i=3;i<=n;i++)
    {
        while (bot<top&&JudgeOut(inter(q[top-1],q[top]),l[i])) top--;
        while (bot<top&&JudgeOut(inter(q[bot],q[bot+1]),l[i])) bot++;
        q[++top]=l[i];
    }
    while (bot<top&&JudgeOut(inter(q[top-1],q[top]),q[bot])) top--;
    while (bot<top&&JudgeOut(inter(q[bot],q[bot+1]),q[top])) bot++;

    return top-bot>1;
}

int main()
{
    zero.x=zero.y=0.0;

    while(scanf("%d",&n)&&n)
    {
        for(int i=1;i<=n;i++)
            scanf("%lf%lf",&p[i].x,&p[i].y);
        p[n+1]=p[1];

        double lft=0.0,rht=10000.0;
        while (rht-lft>eps)
        {
            double mid=(lft+rht)/2.0;
            for(int i=1;i<=n;i++)
            {
                l[i].a.x=p[i].y-p[i+1].y,l[i].a.y=p[i+1].x-p[i].x;
                double r=mid/dis(zero,l[i].a);
                l[i].a.x*=r,l[i].a.y*=r;
                l[i].a.x+=p[i].x,l[i].a.y+=p[i].y;
                l[i].b.x=l[i].a.x+p[i+1].x-p[i].x;
                l[i].b.y=l[i].a.y+p[i+1].y-p[i].y;
            }
            if (solve()) lft=mid;
            else rht=mid;
        }

        printf("%.6f\n",lft);
    }

    return 0;
}