poj 3525 凸多边形多大内切圆
| Time Limit: 5000MS | Memory Limit: 65536K | |||
| Total Submissions: 4758 | Accepted: 2178 | 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 | |
| ⋮ | ||
| xn | yn |
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 (xi, yi)–(xi+1, yi+1) (1 ≤ i ≤n − 1) and the line segment (xn, yn)–(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
4 0 0 10000 0 10000 10000 0 10000 3 0 0 10000 0 7000 1000 6 0 40 100 20 250 40 250 70 100 90 0 70 3 0 0 10000 10000 5000 5001 0
Sample Output
5000.000000 494.233641 34.542948 0.353553
/*
poj 3525 凸多边形多大内切圆
给你一个凸多边形的小岛地图,求出岛中到海岸线的最远距离。
相当于求多边形中最大的内切圆的半径,但是想了很久并没有发现什么方法 卒
问题可以转化成能够在多边形中找到一个圆。也就是二分多边形缩小的长度,然后
判断当前能否找到一个圆。
如果存在一个圆的话,那么它的圆心肯定是这个凸多边形的核。所以可以通过二分
和半平面相交判断解决
hhh-2016-05-11 22:10:56
*/
#include <iostream>
#include <vector>
#include <cstring>
#include <string>
#include <cstdio>
#include <queue>
#include <cmath>
#include <algorithm>
#include <functional>
#include <map>
using namespace std;
#define lson (i<<1)
#define rson ((i<<1)|1)
typedef long long ll;
using namespace std;
const int maxn = 1510;
const double PI = 3.1415926;
const double eps = 1e-8;
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,t;
double k;
Line() {}
Line(Point _s,Point _t)
{
s = _s;
t = _t;
k = atan2(t.y-s.y,t.x-s.x);
}
Point operator &(const Line &b) const
{
Point res = s;
double ta = ((s-b.s)^(b.s-b.t))/((s-t)^(b.s-b.t));
res.x += (t.x-s.x)*ta;
res.y += (t.y-s.y)*ta;
return res;
}
};
bool HPIcmp(Line a,Line b)
{
if(fabs(a.k-b.k) > eps) return a.k<b.k;
return ((a.s-b.s)^(b.t-b.s)) < 0;
}
Line li[maxn];
double CalArea(Point p[],int n)
{
double ans = 0;
for(int i = 0; i < n; i++)
{
ans += (p[i]^p[(i+1)%n])/2;
}
return ans;
}
void HPI(Line line[],int n,Point res[],int &resn)
{
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;
li[0] = line[0];
li[1] = line[1];
resn = 0;
for(int i = 2; i < tot; i++)
{
if(fabs((li[tail].t-li[tail].s)^(li[tail-1].t-li[tail-1].s)) < eps||
fabs((li[head].t-li[head].s)^(li[head+1].t-li[head+1].s)) < eps)
return;
while(head < tail && (((li[tail] & li[tail-1]) - line[i].s) ^ (line[i].t-line[i].s)) > eps)
tail--;
while(head < tail && (((li[head] & li[head+1]) - line[i].s) ^ (line[i].t-line[i].s)) > eps)
head++;
li[++tail] = line[i];
}
while(head < tail && (((li[tail] & li[tail-1]) - li[head].s) ^ (li[head].t-li[head].s)) > eps)
tail--;
while(head < tail && (((li[head] & li[head-1]) - li[tail].s) ^ (li[tail].t-li[tail].t)) > eps)
head++;
if(tail <= head+1)
return;
for(int i = head; i < tail; i++)
res[resn++] = li[i]&li[i+1];
if(head < tail-1)
res[resn++] = li[head]&li[tail];
}
Point p0;
Point lis[maxn];
Line line[maxn];
double dist(Point a,Point b)
{
return sqrt((a-b)*(a-b));
}
bool cmp(Point a,Point b)
{
double t = (a-p0)^(b-p0);
if(sgn(t) > 0)return true;
else if(sgn(t) == 0 && sgn(dist(a,lis[0])-dist(b,lis[0])) <= 0)
return true;
else
return false;
}
Point ta,tb;
Point fans[maxn];
void fin(Point a,Point b,double mid)
{
double len = dist(a,b);
double dx = (a.y-b.y)*mid/len;
double dy = (b.x-a.x)*mid/len;
ta.x = a.x+dx,ta.y = a.y+dy;
tb.x = b.x+dx,tb.y = b.y+dy;
}
int main()
{
//freopen("in.txt","r",stdin);
int n,T;
while(scanf("%d",&n)!= EOF && n)
{
for(int i = 0; i < n; i++)
{
scanf("%lf%lf",&lis[i].x,&lis[i].y);
}
int ans;
double l=0,r=100000;
double tans = 0;
while(r - l > eps)
{
double mid = (l+r) /2;
for(int i = 0; i < n; i++)
{
fin(lis[i],lis[(i+1)%n],mid);
line[i] = Line(ta,tb);
}
HPI(line,n,fans,ans);
if(ans)
l = mid+eps,tans = mid;
else
r = mid-eps;
}
printf("%.6f\n",tans);
}
return 0;
}
