Description
You are given N circles and expected to calculate the area of the union of the circles !
Input
The first line is one integer n indicates the number of the circles. (1 <= n <= 1000)
Then follows n lines every line has three integers
Xi Yi Ri
indicates the coordinate of the center of the circle, and the radius. (|Xi|. |Yi| <= 1000, Ri <= 1000)
Note that in this problem Ri may be 0 and it just means one point !
Output
The total area that these N circles with 3 digits after decimal point
题目大意：求n个圆覆盖的总面积。
思路：参考http://hi.baidu.com/aekdycoin/item/b8ff6adc73c0e71dd78ed0d6
时间复杂度O(n^2*log(n))
代码（0.04S）：
1 #include <cstdio> 2 #include <algorithm> 3 #include <cstring> 4 #include <iostream> 5 #include <cmath> 6 #include <vector> 7 using namespace std; 8 typedef long long LL; 9 10 const double PI = acos(-1.0); 11 const double EPS = 1e-8; 12 13 inline int sgn(double x) { 14 return (x > EPS) - (x < -EPS); 15 } 16 17 struct Point { 18 double x, y; 19 Point() {} 20 Point(double x, double y): x(x), y(y) {} 21 void read() { 22 scanf("%lf%lf", &x, &y); 23 } 24 double angle() { 25 return atan2(y, x); 26 } 27 Point operator + (const Point &rhs) const { 28 return Point(x + rhs.x, y + rhs.y); 29 } 30 Point operator - (const Point &rhs) const { 31 return Point(x - rhs.x, y - rhs.y); 32 } 33 Point operator * (double t) const { 34 return Point(x * t, y * t); 35 } 36 Point operator / (double t) const { 37 return Point(x / t, y / t); 38 } 39 double length() const { 40 return sqrt(x * x + y * y); 41 } 42 Point unit() const { 43 double l = length(); 44 return Point(x / l, y / l); 45 } 46 }; 47 48 double cross(const Point &a, const Point &b) { 49 return a.x * b.y - a.y * b.x; 50 } 51 52 double dist(const Point &p1, const Point &p2) { 53 return (p1 - p2).length(); 54 } 55 56 Point rotate(const Point &p, double angle, const Point &o = Point(0, 0)) { 57 Point t = p - o; 58 double x = t.x * cos(angle) - t.y * sin(angle); 59 double y = t.y * cos(angle) + t.x * sin(angle); 60 return Point(x, y) + o; 61 } 62 63 struct Region { 64 double st, ed; 65 Region() {} 66 Region(double st, double ed): st(st), ed(ed) {} 67 bool operator < (const Region &rhs) const { 68 if(sgn(st - rhs.st)) return st < rhs.st; 69 return ed < rhs.ed; 70 } 71 }; 72 73 struct Circle { 74 Point c; 75 double r; 76 vector<Region> reg; 77 Circle() {} 78 Circle(Point c, double r): c(c), r(r) {} 79 void read() { 80 c.read(); 81 scanf("%lf", &r); 82 } 83 void add(const Region &r) { 84 reg.push_back(r); 85 } 86 bool contain(const Circle &cir) const { 87 return sgn(dist(cir.c, c) + cir.r - r) <= 0; 88 } 89 bool intersect(const Circle &cir) const { 90 return sgn(dist(cir.c, c) - cir.r - r) < 0; 91 } 92 }; 93 94 double sqr(double x) { 95 return x * x; 96 } 97 98 void intersection(const Circle &cir1, const Circle &cir2, Point &p1, Point &p2) { 99 double l = dist(cir1.c, cir2.c); 100 double d = (sqr(l) - sqr(cir2.r) + sqr(cir1.r)) / (2 * l); 101 double d2 = sqrt(sqr(cir1.r) - sqr(d)); 102 Point mid = cir1.c + (cir2.c - cir1.c).unit() * d; 103 Point v = rotate(cir2.c - cir1.c, PI / 2).unit() * d2; 104 p1 = mid + v, p2 = mid - v; 105 } 106 107 Point calc(const Circle &cir, double angle) { 108 Point p = Point(cir.c.x + cir.r, cir.c.y); 109 return rotate(p, angle, cir.c); 110 } 111 112 const int MAXN = 1010; 113 114 Circle cir[MAXN]; 115 bool del[MAXN]; 116 int n; 117 118 double solve() { 119 double ans = 0; 120 for(int i = 0; i < n; ++i) { 121 for(int j = 0; j < n; ++j) if(!del[j]) { 122 if(i == j) continue; 123 if(cir[j].contain(cir[i])) { 124 del[i] = true; 125 break; 126 } 127 } 128 } 129 for(int i = 0; i < n; ++i) if(!del[i]) { 130 Circle &mc = cir[i]; 131 Point p1, p2; 132 bool flag = false; 133 for(int j = 0; j < n; ++j) if(!del[j]) { 134 if(i == j) continue; 135 if(!mc.intersect(cir[j])) continue; 136 flag = true; 137 intersection(mc, cir[j], p1, p2); 138 double rs = (p2 - mc.c).angle(), rt = (p1 - mc.c).angle(); 139 if(sgn(rs) < 0) rs += 2 * PI; 140 if(sgn(rt) < 0) rt += 2 * PI; 141 if(sgn(rs - rt) > 0) mc.add(Region(rs, PI * 2)), mc.add(Region(0, rt)); 142 else mc.add(Region(rs, rt)); 143 } 144 if(!flag) { 145 ans += PI * sqr(mc.r); 146 continue; 147 } 148 sort(mc.reg.begin(), mc.reg.end()); 149 int cnt = 1; 150 for(int j = 1; j < int(mc.reg.size()); ++j) { 151 if(sgn(mc.reg[cnt - 1].ed - mc.reg[j].st) >= 0) { 152 mc.reg[cnt - 1].ed = max(mc.reg[cnt - 1].ed, mc.reg[j].ed); 153 } else mc.reg[cnt++] = mc.reg[j]; 154 } 155 mc.add(Region()); 156 mc.reg[cnt] = mc.reg[0]; 157 for(int j = 0; j < cnt; ++j) { 158 p1 = calc(mc, mc.reg[j].ed); 159 p2 = calc(mc, mc.reg[j + 1].st); 160 ans += cross(p1, p2) / 2; 161 double angle = mc.reg[j + 1].st - mc.reg[j].ed; 162 if(sgn(angle) < 0) angle += 2 * PI; 163 ans += 0.5 * sqr(mc.r) * (angle - sin(angle)); 164 } 165 } 166 return ans; 167 } 168 169 int main() { 170 scanf("%d", &n); 171 for(int i = 0; i < n; ++i) cir[i].read(); 172 printf("%.3f\n", solve() + EPS); 173 }
以下转自：http://hi.baidu.com/aekdycoin/item/b8ff6adc73c0e71dd78ed0d6