POJ 3525(计算几何+凸多边形最大内切圆)

问题描述:

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

题目题意:题目给我们一个凸多边形，让我们求出最大内切圆的半径。

题目分析:凸多边形可以看出凸多边形的边（直线）的半平面的交，其实只要这个半平面的交存在，那么这个内切圆就存在，当半平面的交不存在的时候，内切圆就不存在了。（最后变成一个点)

因为是凸多边形的原因，我们可以通过判断凸多边形的核是否存在来判断。

我们平移凸多边形的各个边，直到核不存在了，平移的距离就是最大内切圆的半径。

代码如下:（半平面交的代码是模板代码)

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;const double eps=1e-8;
const double inf=1e9;
const int MAXN=210;
int m;//保存多边形的点数
double r;//保存内移距离
int cCnt,curCnt;//此时cCnt为最终切割得到的多边形的顶点数、暂存顶点个数
struct point
{double x,y;
};
point points[MAXN],p[MAXN],q[MAXN];//读入的多边形的顶点（顺时针）、p为存放最终切割得到的多边形顶点的数组、暂存核的顶点
void getline(point x,point y,double &a,double &b,double   &c) //两点x、y确定一条直线a、b、c为其系数
{a = y.y - x.y;b = x.x - y.x;c = y.x * x.y - x.x * y.y;
}
void initial()
{for(int i = 1; i <= m; ++i)p[i] = points[i];p[m+1] = p[1];p[0] = p[m];cCnt = m;
}
point intersect(point x,point y,double a,double b,double c)
{double u = fabs(a * x.x + b * x.y + c);double v = fabs(a * y.x + b * y.y + c);point pt;pt.x=(x.x * v + y.x * u) / (u + v);pt.y=(x.y * v + y.y * u) / (u + v);return  pt;
}
void cut(double a,double b ,double c)
{curCnt = 0;for(int i = 1; i <= cCnt; ++i){if(a*p[i].x + b*p[i].y + c >= 0)q[++curCnt] = p[i];else{if(a*p[i-1].x + b*p[i-1].y + c > 0){q[++curCnt] = intersect(p[i],p[i-1],a,b,c);}if(a*p[i+1].x + b*p[i+1].y + c > 0){q[++curCnt] = intersect(p[i],p[i+1],a,b,c);}}}for(int i = 1; i <= curCnt; ++i)p[i] = q[i];p[curCnt+1] = q[1];p[0] = p[curCnt];cCnt = curCnt;
}
int dcmp(double x)
{if(fabs(x)<eps) return 0;else return x<0?-1:1;
}
void solve()
{initial();for(int i = 1; i <= m; ++i){point ta, tb, tt;tt.x = points[i+1].y - points[i].y;tt.y = points[i].x - points[i+1].x;double k = r / sqrt(tt.x * tt.x + tt.y * tt.y);tt.x = tt.x * k;tt.y = tt.y * k;ta.x = points[i].x + tt.x;ta.y = points[i].y + tt.y;tb.x = points[i+1].x + tt.x;tb.y = points[i+1].y + tt.y;double a,b,c;getline(ta,tb,a,b,c);cut(a,b,c);}
}
void GuiZhengHua(){ //规整化方向，逆时针变顺时针，顺时针变逆时针for(int i = 1; i < (m+1)/2; i ++)swap(points[i], points[m-i]);
}
int main()
{while (scanf("%d",&m)!=EOF){if (m==0) break;for (int i=1;i<=m;i++)scanf("%lf%lf",&points[i].x,&points[i].y);GuiZhengHua();points[m+1]=points[1];double left=0,right=inf,mid;while ((right-left)>=eps) {//二分求半径mid=(left+right)/2.0;// cout<<1<<endl;r=mid;solve();if (cCnt<=0) right=mid;else left=mid;}printf("%.6f\n",left);}return 0;
}

POJ 3525(计算几何+凸多边形最大内切圆)相关推荐

POJ 3525/UVA 1396 Most Distant Point from the Sea（二分+半平面交）
Description The main land of Japan called Honshu is an island surrounded by the sea. In such an isla ...
[凸多边形最大内切圆][半平面交]Most Distant Point from the Sea POJ3525
The main land of Japan called Honshu is an island surrounded by the sea. In such an island, it is na ...
poj 3525 tomo带鞘
http://poj.org/problem?id=3525 转载于:https://www.cnblogs.com/ccccnzb/p/3945707.html
[Poj 2187] 计算几何之凸包(二) {更高效的算法}
{ 承上一节继续介绍点集的凸包 (下文中所有凸包若不做特殊说明均指点集的凸包) 这一节介绍相比更高效的算法 } ========================================= ...
POJ 2954-Triangle(计算几何+皮克定理)
题目地址:POJ 2954 题意:给出三角形的三个顶点,求内部格点的个数. 思路:形同POJ 1265. #include <stdio.h> #include <math.h> ...
POJ 1265-Area(计算几何+皮克定理+多边形面积公式)
题目地址:POJ 1265 题意:给定一个格点多边形,求出内部点数in,边上点数on,和面积S. 思路:运用的定理很多. 1.皮克定理:S=in+on/2-1,即in=(2*S+2-on)/2. 2. ...
POJ - Euclid(计算几何)
题目链接:http://poj.org/problem?id=3813 Time Limit: 1000MS Memory Limit: 65536K Description In one of hi ...
POJ 1584 计算几何凸包
链接: http://poj.org/problem?id=1584 题意: 按照顺时针或逆时针方向输入一个n边形的顶点坐标集,先判断这个n边形是否为凸包. 再给定一个圆形(圆心坐标和半径),判断这个 ...
poj 1269 计算几何
题意:此题要处理给定的四个点,前两个点,后两个点分别构成一条直线,判断这两条直线是否重合,平行,相交,若相交,给出交点. 解题思路:可以用向量叉积判断是否重合和平行重合:设第一个点和第二个点构成向量 ...

POJ 3525(计算几何+凸多边形最大内切圆)

POJ 3525(计算几何+凸多边形最大内切圆)相关推荐

最新文章

热门文章