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),判断这个圆是否完全在n变形内部。
题解:
1、判断凸包convex():
由于点集已经按某个时针方向有序,因此可以先定义一个方向系数dir=0
两两枚举n边形的边,用叉积判断这两条边的转向(右螺旋或左螺旋),由于存在散点共线的情况,因此当且仅当叉积的值t第一次不为0时,dir=t,dir的值此后不再改变。(dir>0 则为右螺旋逆时针,dir<0则为左螺旋顺时针)
此后继续枚举剩下的边,只要判断dir*t>=0即可,当存在一个dir*t<0的边,说明这是凹多边形,就不是凸包了。
2、判断圆心在不在凸包内contain()
3、当圆心在凸包内时,判断距离是否都大于半径fit()
代码:
1 #include <map> 2 #include <set> 3 #include <cmath> 4 #include <queue> 5 #include <stack> 6 #include <cstdio> 7 #include <string> 8 #include <vector> 9 #include <cstdlib> 10 #include <cstring> 11 #include <sstream> 12 #include <iostream> 13 #include <algorithm> 14 #include <functional> 15 using namespace std; 16 #define rep(i,a,n) for (int i=a;i<n;i++) 17 #define per(i,a,n) for (int i=n-1;i>=a;i--) 18 #define all(x) (x).begin(),(x).end() 19 #define pb push_back 20 #define mp make_pair 21 #define lson l,m,rt<<1 22 #define rson m+1,r,rt<<1|1 23 typedef long long ll; 24 typedef vector<int> VI; 25 typedef pair<int, int> PII; 26 const ll MOD = 1e9 + 7; 27 const int INF = 0x3f3f3f3f; 28 const int MAXN = 1010; 29 // head 30 31 const double eps = 1e-8; 32 int cmp(double x) { 33 if (fabs(x) < eps) return 0; 34 if (x > 0) return 1; 35 return -1; 36 } 37 38 const double pi = acos(-1); 39 inline double sqr(double x) { 40 return x*x; 41 } 42 struct point { 43 double x, y; 44 point() {} 45 point(double a, double b) :x(a), y(b) {} 46 void input() { 47 scanf("%lf%lf", &x, &y); 48 } 49 friend point operator+(const point &a, const point &b) { 50 return point(a.x + b.x, a.y + b.y); 51 } 52 friend point operator-(const point &a, const point &b) { 53 return point(a.x - b.x, a.y - b.y); 54 } 55 friend point operator*(const double &a, const point &b) { 56 return point(a*b.x, a*b.y); 57 } 58 friend point operator/(const point &a, const double &b) { 59 return point(a.x / b, a.y / b); 60 } 61 double norm() { 62 return sqrt(sqr(x) + sqr(y)); 63 } 64 }; 65 double det(point a, point b) { 66 return a.x*b.y - a.y*b.x; 67 } 68 double dot(point a, point b) { 69 return a.x*b.x + a.y*b.y; 70 } 71 double dist(point a, point b) { 72 return (a - b).norm(); 73 } 74 75 struct line { 76 point a, b; 77 line() {} 78 line(point x, point y) :a(x), b(y) {} 79 }; 80 double dis_point_segment(point p, point s, point t) { 81 if (cmp(dot(p - s, t - s)) < 0) return (p - s).norm(); 82 if (cmp(dot(p - t, s - t)) < 0) return (p - t).norm(); 83 return fabs(det(s - p, t - p) / dist(s, t)); 84 } 85 bool point_on_segment(point p, point s, point t) { 86 return cmp(det(p - s, t - s)) == 0 && cmp(dot(p - s, p - t)) <= 0; 87 } 88 bool parallel(line a, line b) { 89 return !cmp(det(a.a - a.b, b.a - b.b)); 90 } 91 bool line_make_point(line a, line b,point &res) { 92 if (parallel(a, b)) return false; 93 double s1 = det(a.a - b.a, b.b - b.a); 94 double s2 = det(a.b - b.a, b.b - b.a); 95 res = (s1*a.b - s2*a.a) / (s1 - s2); 96 return true; 97 } 98 99 int n; 100 double r; 101 point o; 102 point *p; 103 104 bool convex() { 105 int dir = 0; 106 rep(i, 0, n) { 107 int t = cmp(det(p[i + 1] - p[i], p[i + 2] - p[i + 1])); 108 if (!dir) dir = t; 109 if (dir*t < 0) return false; 110 } 111 return true; 112 } 113 114 bool contain() { 115 int sign = 0; 116 rep(i, 0, n) { 117 int x = cmp(det(p[i] - o, p[i + 1] - o)); 118 if (x) { 119 if (sign && sign != x) return false; 120 else sign = x; 121 } 122 } 123 return true; 124 } 125 126 bool fit() { 127 rep(i, 0, n) { 128 int k = cmp(dis_point_segment(o, p[i], p[i + 1]) - r); 129 if (k < 0) return false; 130 } 131 return true; 132 } 133 134 int main() { 135 while (cin >> n && n != 1) { 136 cin >> r; 137 o.input(); 138 p = new point[n + 2]; 139 rep(i, 1, n + 1) p[i].input(); 140 p[0] = p[n]; 141 p[n + 1] = p[1]; 142 if(!convex()) cout << "HOLE IS ILL-FORMED" << endl; 143 else { 144 bool fg1 = contain(); 145 bool fg2 = fit(); 146 if (fg1 && fg2) cout << "PEG WILL FIT" << endl; 147 else cout << "PEG WILL NOT FIT" << endl; 148 } 149 delete p; 150 } 151 return 0; 152 }