多圆面积交模板

main函数中只要调用solve（）就好了，solve函数中的n即代表n个圆，先输入n，再输入n个圆的圆心坐标和半径，直接在函数中输出多圆相交的公共面积

main函数中不用做什么，所有的东西都在solve完成了

 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);
     //去掉重复计算的
     for (int i = 1; i <= n; ++i)
     {
         area[i] -= area[i + 1];
     }
     //area[i]为重叠了i次的面积
     //tot 为总面积
     double tot = 0;
     for(int i=1; i<=n; i++) tot += area[i];
     printf("%f\n", tot);
 }

 int main()
 {
     //freopen("input.txt", "r", stdin);
     return 0;
 }