POJ 3525 半平面交
题意:
求凸包内切圆最大半径
题解:
二分半径,将凸包所有边往凸包内平移这么半径长度,看平移后是否能围成凸包.
/*
* File: main.cpp
* Author: swordholy
*
* Created on 2011年3月25日, 下午7:56
*/
//求凸包内切圆最大半径
#include <cstdlib>
#include <iostream>
#include <stdio.h>
#include <memory.h>
#include <algorithm>
#include <cmath>
using namespace std;
#define MAXN 110
#define eps 1e-10
#define zero(x) (((x)>0?(x):-(x))<eps)
struct pt
{
double x,y;
pt()
{
}
pt(double xx,double yy)
{
x=xx;
y=yy;
}
};
double dist(pt p1,pt p2)
{
return sqrt(1.0*(p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y));
}
double Cross(pt a,pt b)
{
return a.x*b.y-a.y*b.x;
}
double Cross(pt a,pt b,pt c)
{
return Cross(pt(a.x-c.x,a.y-c.y),pt(b.x-c.x,b.y-c.y));
}
double same_side(pt p1,pt p2,pt l1,pt l2)
{
return Cross(l1,p1,l2)*Cross(l1,p2,l2)>eps;
}
struct Convex
{
int n;
pt p[MAXN];
};
struct Plane//半平面,a,b构成直线,side表示在哪个面
{
pt a,b,side;
};
/*
ret.x-u1.x=t(u2.x-u1.x)
ret.y-u1.y=t(u2.y-u1.y)
ret.x-v1.x=tt(v2.x-v1.x)
ret.y-v1.y=tt(v2.y-v1.y)
解以上方程可得线段交点ret
*/
pt intersection(pt u1,pt u2,pt v1,pt v2)
{
pt ret=u1;
double t=((u1.x-v1.x)*(v1.y-v2.y)-(u1.y-v1.y)*(v1.x-v2.x))
/((u1.x-u2.x)*(v1.y-v2.y)-(u1.y-u2.y)*(v1.x-v2.x));
ret.x+=(u2.x-u1.x)*t;
ret.y+=(u2.y-u1.y)*t;
return ret;
}
Convex src,ans;
Convex Half_Plane(Convex c,Plane pl)//半平面交,一条直线切割凸包
{
int i,j;
double r1,r2;
Convex ans,ans1;
ans.n=0;
for(i=0;i<c.n;i++)
{
if (same_side(c.p[i],pl.side,pl.a,pl.b))
ans.p[ans.n++]=c.p[i];
if ((!same_side(c.p[i],c.p[(i+1)%c.n],pl.a,pl.b))&&(!(zero(Cross(c.p[i],pl.a,pl.b))&&zero(Cross(c.p[(i+1)%c.n],pl.a,pl.b)))))
ans.p[ans.n++]=intersection(c.p[i],c.p[(i+1)%c.n],pl.a,pl.b);//有点在直线上的话该点会被加入两次
}
ans1.n=0;
for(i=0;i<ans.n;i++)
if (!i||!zero(ans.p[i].x-ans.p[i-1].x)||!zero(ans.p[i].y-ans.p[i-1].y))
ans1.p[ans1.n++]=ans.p[i];//过滤重复点
if (zero(ans1.p[ans1.n-1].x-ans1.p[0].x)&&zero(ans1.p[ans1.n-1].y-ans1.p[0].y))
ans1.n--;
if (ans1.n<3)
ans1.n=0;
return ans1;
}
bool judge(double x)
{
Convex ans;
pt tt,ta,tb;
ans=src;
for (int i = 0; i < src.n; i++)
{
tt.x = src.p[i].y - src.p[i+1].y;
tt.y = src.p[i + 1].x - src.p[i].x;
double k = x / sqrt(tt.x * tt.x + tt.y * tt.y);
tt.x = tt.x*k;
tt.y = tt.y*k;
ta.x = src.p[i].x + tt.x;
ta.y = src.p[i].y + tt.y;
tb.x = src.p[i + 1].x + tt.x;
tb.y = src.p[i + 1].y + tt.y;
Plane pl;
pl.a=ta;pl.b=tb;
pl.side.x = src.p[i].x + tt.x + tt.x;
pl.side.y = src.p[i].y + tt.y + tt.y;
ans=Half_Plane(ans,pl);
if (ans.n==0) return false;
}
return true;
}
int main(int argc, char** argv)
{
int n,i,j,tcase;
double maxl;
while(scanf("%d",&n)!=EOF)
{
src.n=n;
for(i=0;i<n;i++)
scanf("%lf %lf",&src.p[i].x,&src.p[i].y);
src.p[n]=src.p[0];
double d,maxd=0;
for(i=0;i<n;i++)
for(j=0;j<n;j++)
if ((d=dist(src.p[i],src.p[j]))>maxd)
maxd=d;
double l,r,mid,k,ansd;
l=0;r=maxd/2;
ansd=0;
while(l+eps<r)
{
mid=(l+r)/2;
if ( judge(mid) )
{
ansd=mid;
l=mid+eps;
}
else
r=mid-eps;
}
printf("%.6lf/n",ansd);
}
return 0;
}