多圓相併面積
輸入n代表n個圓，輸入n個圓的圓心座標和半徑，函式輸出多圓相併的公共面積

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef vector <int> VI;
const int INF = 0x3f3f3f3f;
const double eps = 1e-10;
const int MOD = 100000007;
const int MAXN = 1000010;
const double PI = acos(-1.0);
#define sqr(x) ((x)*(x))
const int N = 1010;
double area[N];
int n;
int dcmp(double x){
    if (x < -eps) return -1;
    else return x > eps;
}
struct cp{
    double x, y, r, angle;
    int d;
    cp() {}
    cp(double xx, double yy, double ang = 0, int t = 0){
        x = xx;
        y = yy;
        angle = ang;
        d = t;
    }
    void get(){
        scanf("%lf%lf%lf", &x, &y, &r);
        d = 1;
    }
} cir[N], tp[N * 2];
double dis(cp a, cp b){return sqrt(sqr(a.x - b.x) + sqr(a.y - b.y));}
double cross(cp p0, cp p1, cp p2){return (p1.x - p0.x) * (p2.y - p0.y) - (p1.y - p0.y) * (p2.x - p0.x);}
int CirCrossCir(cp p1, double r1, cp p2, double r2, cp &cp1, cp &cp2){
    double mx = p2.x - p1.x, sx = p2.x + p1.x, mx2 = mx * mx;
    double my = p2.y - p1.y, sy = p2.y + p1.y, my2 = my * my;
    double sq = mx2 + my2, d = -(sq - sqr(r1 - r2)) * (sq - sqr(r1 + r2));
    if (d + eps < 0) return 0;
    if (d < eps) d = 0;
    else d = sqrt(d);
    double x = mx * ((r1 + r2) * (r1 - r2) + mx * sx) + sx * my2;
    double y = my * ((r1 + r2) * (r1 - r2) + my * sy) + sy * mx2;
    double dx = mx * d, dy = my * d;
    sq *= 2;
    cp1.x = (x - dy) / sq;
    cp1.y = (y + dx) / sq;
    cp2.x = (x + dy) / sq;
    cp2.y = (y - dx) / sq;
    if (d > eps) return 2;
    else return 1;
}
bool circmp(const cp& u, const cp& v){return dcmp(u.r - v.r) < 0;}
bool cmp(const cp& u, const cp& v){
    if (dcmp(u.angle - v.angle)) return u.angle < v.angle;
    return u.d > v.d;
}
double calc(cp cir, cp cp1, cp cp2){
    double ans = (cp2.angle - cp1.angle) * sqr(cir.r)
                 - cross(cir, cp1, cp2) + cross(cp(0, 0), cp1, cp2);
    return ans / 2;
}
void CirUnion(cp cir[], int n){
    cp cp1, cp2;
    sort(cir, cir + n, circmp);
    for (int i = 0; i < n; ++i)
        for (int j = i + 1; j < n; ++j)
            if (dcmp(dis(cir[i], cir[j]) + cir[i].r - cir[j].r) <= 0)
                cir[i].d++;
    for (int i = 0; i < n; ++i){
        int tn = 0, cnt = 0;
        for (int j = 0; j < n; ++j){
            if (i == j) continue;
            if (CirCrossCir(cir[i], cir[i].r, cir[j], cir[j].r,cp2, cp1) < 2) continue;
            cp1.angle = atan2(cp1.y - cir[i].y, cp1.x - cir[i].x);
            cp2.angle = atan2(cp2.y - cir[i].y, cp2.x - cir[i].x);
            cp1.d = 1;
            tp[tn++] = cp1;
            cp2.d = -1;
            tp[tn++] = cp2;
            if (dcmp(cp1.angle - cp2.angle) > 0) cnt++;
        }
        tp[tn++] = cp(cir[i].x - cir[i].r, cir[i].y, PI, -cnt);
        tp[tn++] = cp(cir[i].x - cir[i].r, cir[i].y, -PI, cnt);
        sort(tp, tp + tn, cmp);
        int p, s = cir[i].d + tp[0].d;
        for (int j = 1; j < tn; ++j){
            p = s;
            s += tp[j].d;
            area[p] += calc(cir[i], tp[j - 1], tp[j]);
        }
    }
}
void solve(){
    scanf("%d", &n);
    for (int i = 0; i < n; ++i)cir[i].get();
    memset(area, 0, sizeof(area));
    CirUnion(cir, n);
    for (int i = 1; i <= n; ++i)    //去掉重複計算的
        area[i] -= area[i + 1];     //area[i]為重疊了i次的面積
    double tot = 0;//tot 為總面積
    for(int i=1; i<=n; i++) tot += area[i];
    cout<<area[n]<<endl;    //重疊了n次就是所求N圓相交面積
    //printf("%f\n", tot);  //這句就是求總面積
}
int main(){
    solve();
    return 0;
}
/*
in
3
1 0 2
-1 0 2
0 1 2
Out
4.20324 相交面積
//22.929505 相併面積
*/