题解 SP8073 【CIRU - The area of the union of circles】

题目链接

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

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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 << '\n';
    return 0;
}

来源：https://www.cnblogs.com/colazcy/p/12061397.html