倒过来做，然后就变成了线段覆盖问题了。
线段覆盖就是贪心即可。。。
但是好烦好烦= =，需要耐心和几何基础2333
1 /************************************************************** 2 Problem: 1043 3 User: rausen 4 Language: C++ 5 Result: Accepted 6 Time:240 ms 7 Memory:872 kb 8 ****************************************************************/ 9 10 #include <cstdio> 11 #include <cmath> 12 #include <algorithm> 13 14 using namespace std; 15 typedef double lf; 16 typedef pair <lf, lf> Pair; 17 const lf pi = 3.1415926536; 18 const int N = 1005; 19 20 struct point { 21 lf x, y; 22 }; 23 typedef point P; 24 25 struct circle { 26 P o; 27 lf r; 28 }a[N]; 29 typedef circle C; 30 31 int n, tot; 32 lf ans; 33 Pair inter[N << 1]; 34 35 inline lf Sqr(lf x) { 36 return x * x; 37 } 38 39 inline lf dist(const P & a, const P &b) { 40 return sqrt(Sqr(a.x - b.x) + Sqr(a.y - b.y)); 41 } 42 43 inline lf K(const P &a, const P &b) { 44 return atan2(a.y - b.y, a.x - b.x); 45 } 46 47 void Calc (C O1, C O2, lf &dis) { 48 lf alpha = K(O1.o, O2.o) + pi, 49 delta = acos((Sqr(O1.r) + Sqr(dis) - Sqr(O2.r)) / (2 * O1.r * dis)); 50 Pair tmp = make_pair(alpha - delta, alpha + delta); 51 if (tmp.first >= 0 && tmp.second <= 2 * pi) 52 inter[++tot] = tmp; 53 else if (tmp.first < 0) 54 inter[++tot] = make_pair(tmp.first + 2 * pi, 2 * pi), 55 inter[++tot] = make_pair(0, tmp.second); 56 else 57 inter[++tot] = make_pair(tmp.first, 2 * pi), 58 inter[++tot] = make_pair(0, tmp.second - 2 * pi); 59 } 60 61 lf Union() { 62 int i; 63 lf res = 0, st = -1, ed = -1; 64 sort(inter + 1, inter + tot + 1); 65 for (i = 1; i <= tot; ++i) { 66 if (inter[i].first > ed) { 67 res += ed - st; 68 st = inter[i].first, ed = inter[i].second; 69 } else 70 ed = max(ed, inter[i].second); 71 } 72 res += ed - st; 73 return 2 * pi - res; 74 } 75 76 int main() { 77 int i, j; 78 lf dis; 79 scanf("%d ", &n); 80 for (i = n; i; --i) 81 scanf("%lf%lf%lf", &a[i].r, &a[i].o.x, &a[i].o.y); 82 for (i = 1; i <= n; ++i) { 83 tot = 0; 84 for (j = 1; j < i; ++j) { 85 dis = dist(a[i].o, a[j].o); 86 if (a[j].r - a[i].r > dis) break; 87 if (a[i].r + a[j].r > dis && fabs(a[i].r - a[j].r) < dis) 88 Calc(a[i], a[j], dis); 89 } 90 if (j == i) 91 ans += Union() * a[i].r; 92 } 93 printf("%.3lf ", ans); 94 return 0; 95 }