hdu的1558题，为何这样写代码，会memory limit exceed

Question

#include<stdio.h>
#include<cmath>
using namespace std;

struct Point {
    double x, y;
    Point(double X = 0, double Y = 0) { x = X, y = Y; }
    Point operator + (Point B) { return Point(x + B.x, y + B.y); }
    Point operator - (Point B) { return Point(x - B.x, y - B.y); }
    Point operator * (double k) { return Point(x*k, y*k); }
    Point operator / (double k) { return Point(x / k, y / k); }
};
struct Line {
    Point p1, p2;
    Line() {}
    Line(Point P1, Point P2) { p1 = P1; p2 = P2; }
};

typedef Point Vector;

const double eps = 1e-8;
Line line[100];
int pre[100];

int sgn(double a) {
    if (fabs(a) < eps)
        return 0;
    else
        return a < 0 ? -1 : 1;
}

double Cross(Vector A, Vector B) { return B.y*A.x - B.x*A.y; }

bool Cross_segment(Point a, Point b, Point c, Point d) {    //判断两线段是否相交
    double c1 = Cross(b - a, c - a), c2 = Cross(b - a, d - a);
    double d1 = Cross(d - c, a - c), d2 = Cross(d - c, b - c);
    return sgn(c1)*sgn(c2) < 0 && sgn(d1)*sgn(d2) < 0;  //1相交 0不相交 
}

int find(int u) {       //并查集搜索
    return pre[u] == u ? u : pre[u] = find(pre[u]);
}

int main() {
    int ccase;
    scanf("%d", &ccase);
    while (ccase--) {
        for (int i = 0; i < 100; i++)
            pre[i] = i;

        int k = 1;  //线段数 
        char t;
        int n;
        scanf("%d", &n);        //指令数 
        while (n--) {
            t = getchar();
            while (t == ' ' || t == '\n')
                t = getchar();
            if (t == 'P') {
                scanf("%lf%lf%lf%lf", &line[k].p1.x, &line[k].p1.y, &line[k].p2.x, &line[k].p2.y);
                k++;
                for (int i = 1; i < k - 1; i++) {
                    if (Cross_segment(line[k - 1].p1, line[k - 1].p2, line[i].p1, line[i].p2)) {        //如果两线段相交
                        int x = pre[k - 1];
                        int y = pre[i];
                        if (x != y)
                            pre[x] = y;
                    }
                }
            }
            else {
                int x;
                int num = 0;
                scanf("%d", &x);
                x = find(x);
                for (int i = 1; i < k; i++)
                    if (find(i) == x)
                        num++;
                printf("%d\n", num);
            }

        }
        printf("\n");
    }

    return 0;
}

Answer 1

int find(int u) { //并查集搜索
return pre[u] == u ? u : pre[u] = find(pre[u]);
}
是不是这里无限递归了