poj1584A Round Peg in a Ground Hole

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题意甚是难懂！这是第二遍做这道题了，依旧无法理解题意，搜了下题意。。。

首先需要判断是不是为凸多边形。（从一个顶点走一遍即可，要注意顺逆时针，题目中没有指明）

其次看一下圆是不是能够放入多边形内。（首先判断一下圆心是否在圆内，然后枚举圆心到所有边的距离与半径r进行比较）

  1 #include <iostream>
  2 #include<cstdio>
  3 #include<cstring>
  4 #include<algorithm>
  5 #include<stdlib.h>
  6 #include<vector>
  7 #include<cmath>
  8 #include<queue>
  9 #include<set>
 10 using namespace std;
 11 #define N 10000
 12 #define LL long long
 13 #define INF 0xfffffff
 14 const double eps = 1e-8;
 15 const double pi = acos(-1.0);
 16 const double inf = ~0u>>2;
 17 #define _sign(x) ((x) > eps?1:((x) < -eps ? 2 :0))
 18 struct Point
 19 {
 20     double x,y;
 21     Point(double x=0,double y=0):x(x),y(y) {}
 22 }p[N],ch[N];
 23 int top;
 24 struct Circle
 25 {
 26     Point c;
 27     double r;
 28     //Circle(Point c,double r):c(c),r(r){}
 29     Point point (double a)
 30     {
 31         return Point(c.x+cos(a)*r,c.y+sin(a)*r);
 32     }
 33 };
 34 typedef Point pointt;
 35 pointt operator + (Point a,Point b)
 36 {
 37     return Point(a.x+b.x,a.y+b.y);
 38 }
 39 pointt operator - (Point a,Point b)
 40 {
 41     return Point(a.x-b.x,a.y-b.y);
 42 }
 43 double cross(Point a,Point b)
 44 {
 45     return a.x*b.y-a.y*b.x;
 46 }
 47 double mul(Point p0,Point p1,Point p2)
 48 {
 49     return cross(p1-p0,p2-p0);
 50 }
 51 double dis(Point a)
 52 {
 53     return sqrt(a.x*a.x+a.y*a.y);
 54 }
 55 //精度判正负
 56 int dcmp(double x)
 57 {
 58     if(fabs(x)<eps) return 0;
 59     else return x<0?-1:1;
 60 }
 61 int Graham(int n)
 62 {
 63     int i,s[3] = {1,1,1};
 64     double t;
 65     for(i = 0; i < n&&s[1]|s[2]; i ++)
 66     {
 67         t = mul(p[(i+1)%n],p[(i+2)%n],p[i]);
 68         s[_sign(t)] = 0;
 69     }
 70     return s[1]|s[2];
 71 }
 72 double distoline(Point p,Point a,Point b)
 73 {
 74     Point v1 = b-a,v2 = p-a;
 75     return fabs(cross(v1,v2)/dis(v1));
 76 }
 77 int inside(Point po,int n)
 78 {
 79     int i,s[3] = {1,1,1};
 80     double t;
 81     for(i = 0;i < n&&s[1]|s[2];i ++)
 82     {
 83         t = mul(p[(i+1)%n],po,p[i]);
 84         s[_sign(t)] = 0;
 85     }
 86     return s[1]|s[2];
 87 }
 88 int main()
 89 {
 90     int n,i;
 91     Circle cc;
 92     while(scanf("%d",&n)!=EOF)
 93     {
 94         if(n<3) break;
 95         scanf("%lf%lf%lf",&cc.r,&cc.c.x,&cc.c.y);
 96         for(i  = 0; i < n; i++)
 97         scanf("%lf%lf",&p[i].x,&p[i].y);
 98         top = 0;
 99         int m = Graham(n);
100         if(!m)
101         {
102             puts("HOLE IS ILL-FORMED");
103             continue;
104         }
105         int flag = 1;
106         if(!inside(cc.c,n))
107         {
108             puts("PEG WILL NOT FIT");
109             continue;
110         }
111         for(i = 1; i < n; i ++)
112         {
113             if(dcmp(distoline(cc.c,p[i],p[i-1])-cc.r)<0)
114             {
115                 flag = 0;
116                 break;
117             }
118         }
119         if(!flag)
120         puts("PEG WILL NOT FIT");
121         else
122         puts("PEG WILL FIT");
123     }
124     return 0;
125 }
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posted @ 2014-07-16 17:56  _雨  阅读(154)  评论(0)  编辑  收藏  举报