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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