Solution 圆的面积并
题目大意:给定(n)个圆,(n leq 1000)求这些圆的面积并
自适应辛普森
求圆的面积并我们不好求,但是我们可以(O(nlogn))求出(n)条线段的并,于是我们可以用(f(X))表示直线(x = X)与每个圆的切线的并,然后我们求(int_{-infty}^{infty} f(x);dx)即可
直接上自适应辛普森,不过要注意,如果朴素算法及其容易炸精度(比如圆分布的比较稀疏,你辛普森算法第一层取的(5)个点全都是(0)就gg了,可以分段做辛普森)
#include <algorithm>
#include <iostream>
#include <iomanip>
#include <vector>
#include <cmath>
#include <cstdio>
#include <map>
using namespace std;
const int maxn = 1024;
double eps = 1e-4;
struct Seg{
double l,r;
bool operator < (const Seg &rhs)const{return l < rhs.l;}
}seg[maxn];
struct Round{
double x,y,r;
}val[maxn];
int n;
map<double,double> mp;
inline double f(double x){
if(mp[x])return mp[x];
static Seg seg[maxn];
int tot = 0;
for(int i = 1;i <= n;i++){
double deltay = val[i].r * val[i].r - (val[i].x - x) * (val[i].x - x);
if(deltay <= eps)continue;
deltay = sqrt(deltay);
seg[++tot] = Seg{val[i].y - deltay,val[i].y + deltay};
}
sort(seg + 1,seg + 1 + tot);
double res = 0,last = -1e9;
for(int i = 1;i <= tot;i++){
const Seg &now = seg[i];
if(now.l > last)res += now.r - now.l,last = now.r;
else if(now.r > last)res += now.r - last,last = now.r;
}
return mp[x] = res;
}
inline double simpson(double l,double r){
double mid = (l + r) / 2.0;
return (r - l) * (f(l) + f(mid) * 4.0 + f(r)) / 6.0;
}
inline double solve(double l,double r,double now,double eps){
double mid = (l + r) / 2.0,L = simpson(l,mid),R = simpson(mid,r);
if(fabs(L + R - now) < eps)return L + R;
return solve(l,mid,L,eps / 2) + solve(mid,r,R,eps / 2);
}
double l = 1e9,r = -1e9,ans,last = -1e9;
int main(){
ios::sync_with_stdio(false);
cin >> n;
for(int i = 1;i <= n;i++){
cin >> val[i].x >> val[i].y >> val[i].r;
seg[i] = Seg{val[i].x - val[i].r,val[i].x + val[i].r};
l = min(l,val[i].x - val[i].r);
r = max(r,val[i].x + val[i].r);
}
sort(seg + 1,seg+ 1 + n);
for(int i = 1;i <= n;i++){
const Seg &now = seg[i];
if(now.l > last)ans += solve(now.l,now.r,simpson(now.l,now.r),eps),last = now.r;
else if(now.r > last)ans += solve(last,now.r,simpson(now.l,now.r),eps),last = now.r;
}
cout << setiosflags(ios::fixed) << setprecision(3) << ans << '
';
return 0;
}