POJ 1584 A Round Peg in a Ground Hole【计算几何=_=你值得一虐】

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A Round Peg in a Ground Hole
Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 4475   Accepted: 1374

Description

The DIY Furniture company specializes in assemble-it-yourself furniture kits. Typically, the pieces of wood are attached to one another using a wooden peg that fits into pre-cut holes in each piece to be attached. The pegs have a circular cross-section and so are intended to fit inside a round hole. 
A recent factory run of computer desks were flawed when an automatic grinding machine was mis-programmed. The result is an irregularly shaped hole in one piece that, instead of the expected circular shape, is actually an irregular polygon. You need to figure out whether the desks need to be scrapped or if they can be salvaged by filling a part of the hole with a mixture of wood shavings and glue. 
There are two concerns. First, if the hole contains any protrusions (i.e., if there exist any two interior points in the hole that, if connected by a line segment, that segment would cross one or more edges of the hole), then the filled-in-hole would not be structurally sound enough to support the peg under normal stress as the furniture is used. Second, assuming the hole is appropriately shaped, it must be big enough to allow insertion of the peg. Since the hole in this piece of wood must match up with a corresponding hole in other pieces, the precise location where the peg must fit is known. 
Write a program to accept descriptions of pegs and polygonal holes and determine if the hole is ill-formed and, if not, whether the peg will fit at the desired location. Each hole is described as a polygon with vertices (x1, y1), (x2, y2), . . . , (xn, yn). The edges of the polygon are (xi, yi) to (xi+1, yi+1) for i = 1 . . . n − 1 and (xn, yn) to (x1, y1).

Input

Input consists of a series of piece descriptions. Each piece description consists of the following data: 
Line 1 < nVertices > < pegRadius > < pegX > < pegY > 
number of vertices in polygon, n (integer) 
radius of peg (real) 
X and Y position of peg (real) 
n Lines < vertexX > < vertexY > 
On a line for each vertex, listed in order, the X and Y position of vertex The end of input is indicated by a number of polygon vertices less than 3.

Output

For each piece description, print a single line containing the string: 
HOLE IS ILL-FORMED if the hole contains protrusions 
PEG WILL FIT if the hole contains no protrusions and the peg fits in the hole at the indicated position 
PEG WILL NOT FIT if the hole contains no protrusions but the peg will not fit in the hole at the indicated position

Sample Input

5 1.5 1.5 2.0
1.0 1.0
2.0 2.0
1.75 2.0
1.0 3.0
0.0 2.0
5 1.5 1.5 2.0
1.0 1.0
2.0 2.0
1.75 2.5
1.0 3.0
0.0 2.0
1

Sample Output

HOLE IS ILL-FORMED
PEG WILL NOT FIT

Source

Mid-Atlantic 2003



题意：


给你一个含有 N个点的多边形和一个钉子
        判断钉子是否在多边形内部


 注意：

 

输入的第一行先输入多边形的点数,   
            再输入的是钉子的半径,然后才是坐标ToT



思路：


1.先判断多边形是不是凸多边形,
           如果不是,则输出 HOLE IS ILL-FORMED
          如果是,则继续往下判断
       2.(1)判断圆心是否在凸多边形外面
            如果在外面,直接返回 false
            如果在边上,而且半径 == 0,返回 true
                       半径不为 0 , 返回 false
            如果在内部,则遍历圆心到每一条边线段的距离是否 >= 半径
                       如果全部满足,则返回 true
                       否则返回 false
         如果钉子能装下,则输出PEG WILL FIT
         否则输出PEG WILL NOT FIT


忠告：不要NC


相关测试题目：




开始一直WA直到找了这三道基础的题目AC完



/***************************************************
Accepted	192 KB	0 ms	C++	4208 B	2013-07-28 16:02:24
题意：给你一个含有 N个点的多边形和一个钉子
      判断钉子是否在多边形内部
      注意：输入的第一行先输入多边形的点数,
            再输入的是钉子的半径,然后才是坐标ToT

思路：1.先判断多边形是不是凸多边形,
        如果不是,则输出 HOLE IS ILL-FORMED
        如果是,则继续往下判断
      2.(1)判断圆心是否在凸多边形外面
           如果在外面,直接返回 false
           如果在边上,而且半径 == 0,返回 true
                       半径不为 0 , 返回 false
           如果在内部,则遍历圆心到每一条边线段的距离是否 >= 半径
                       如果全部满足,则返回 true
                       否则返回 false
        如果钉子能装下,则输出PEG WILL FIT
        否则输出PEG WILL NOT FIT
***************************************************/
#include<stdio.h>
#include<math.h>

const int maxn = 200;

struct Point{
    double x,y;

    Point() {}
    Point(double _x, double _y) {
        x = _x;
        y = _y;
    }
    Point operator - (const Point &B)
    {
        return Point(x-B.x, y-B.y);
    }
}p[maxn];

struct Circle{
    Point center;
    double radius;
}c;

const double eps = 1e-5;
int dcmp(double x)
{
    if(fabs(x) < 0) return 0;
    else return x < 0 ? -1 : 1;
}

bool operator == (const Point &A, const Point &B)
{
    return dcmp(A.x-B.x)== 0 && dcmp(A.y-B.y) == 0;
}

double Cross(Point A, Point B) /** 叉积*/
{
    return A.x*B.y - A.y*B.x;
}
double Dot(Point A, Point B) /** 点积*/
{
    return A.x*B.x+A.y*B.y;
}

double Length(Point A)
{
    return sqrt(A.x*A.x + A.y*A.y);
}

/** 判断多边形是否是凸多边形【含共线】*/
bool isConvex(Point *p, int n)
{
    p[n] = p[0]; // 边界处理
    p[n+1] = p[1]; // 注意也可以用 %n 处理, 下标从 0 开始
    int now = dcmp(Cross(p[1]-p[0], p[2]-p[1]));
    for(int i = 1; i < n; i++)
    {
        int next = dcmp(Cross(p[i+1]-p[i], p[i+2]-p[i+1]));
        if(now*next < 0) //此处可以共线
        {
            return false;
        }
        now = next; //注意记录临界条件
    }
    return true;
}

/** 点Point 是否在有 n 个顶点的凸多边形内【含边界】*/
int isPointInConvex(Point *p, int n, Point point)
{
    int flag = 1;
    p[n] = p[0];
    int now = dcmp(Cross(p[0]-point, p[1]-point));
    for(int i = 1; i < n; i++)
    {
        int next = dcmp(Cross(p[i]-point, p[i+1]-point));
        if(next*now < 0)
        {
             return -1; /** 点在外面*/
        }
        else if(next*now == 0)
        {
            return 0; /** 点在边上 */
        }
        now = next;
    }
    return flag; /** 点在内部*/
}

/** 判断点P 到线段 AB的距离*/
double DistanceToSegment(Point P, Point A, Point B)
{
    if(A == B) return Length(P-A);
    Point v1 = B-A;
    Point v2 = P-A;
    Point v3 = P-B;

    if(dcmp(Dot(v1, v2)) < 0) return Length(v2);
    else if(dcmp(Dot(v1, v3)) > 0) return Length(v3);
    else return fabs(Cross(v1, v2))/ Length(v1); //忠告：不要脑残的 / 2...
}

/** 判断圆是否在凸多边形内部, 相切也可以*/
bool isCircleInConvex(Point *p, int n, Circle c)
{
    int flag = isPointInConvex(p, n, c.center); /**判断圆心*/

    if(flag == 0) /** 圆心在边上*/
    {
        if(c.radius == 0) return true;
        else return false;
    }
    else if(flag == 1) /** 圆心在内部*/
    {
        p[n] = p[0]; /** 边界处理*/
        for(int i = 0; i < n; i++) /** 遍历所有的边 */
        {
            if(dcmp(DistanceToSegment(c.center, p[i], p[i+1])-c.radius) < 0)
            {
                return false;
            }
        }
        return true;
    }
    else return false; /** 圆心在外部*/
}

int main()
{
    int n;
    while(scanf("%d", &n) != EOF)
    {
        if(n < 3) break;

        /** 忠告：输入时注意顺序, 不要脑残。。。*/
        scanf("%lf%lf%lf", &c.radius, &c.center.x, &c.center.y);
        for(int i = 0; i < n; i++)
            scanf("%lf%lf", &p[i].x, &p[i].y);

        bool flag = isConvex(p, n); /** 判断是否是凸多边形*/
        if(flag)
        {
            flag = isCircleInConvex(p, n, c);
            if(flag) printf("PEG WILL FIT\n");
            else printf("PEG WILL NOT FIT\n");
        }
        else printf("HOLE IS ILL-FORMED\n");
    }
    return 0;
}








posted @ 2013-07-28 16:40  free斩  Views(209)  Comments(0)  Edit  收藏  举报