HUST 1376 Random intersection
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神题。同学指教。1秒AC。。。http://blog.csdn.net/jtjy568805874/article/details/50724656
#include<cstdio> #include<cstring> #include<ctime> #include<algorithm> using namespace std; const int maxn = 100000 + 10; struct point { double x; double y; int id; }p[2 * maxn]; struct Line { int a; int b; }line[maxn]; int T, n, tot; int flag[maxn]; bool cmp(const point&a, const point&b) { if (a.x == b.x) return a.y < b.y; return a.x < b.x; } const double eps = 1e-8; #define zero(x)(((x)>0?(x):(-x))<eps) double xmult(point p1, point p2, point p0) { return (p1.x - p0.x)*(p2.y - p0.y) - (p2.x - p0.x)*(p1.y - p0.y); } int dots_inline(point p1, point p2, point p3) { return zero(xmult(p1, p2, p3)); } int same_side(point p1, point p2, point l1, point l2) { return xmult(l1, p1, l2)*xmult(l1, p2, l2)>eps; } int dot_online_in(point p, point l1, point l2) { return zero(xmult(p, l1, l2)) && (l1.x - p.x)*(l2.x - p.x)<eps && (l1.y - p.y)*(l2.y - p.y)<eps; } int intersect_in(point u1, point u2, point v1, point v2) { if (!dots_inline(u1, u2, v1) || !dots_inline(u1, u2, v2)) return !same_side(u1, u2, v1, v2) && !same_side(v1, v2, u1, u2); return dot_online_in(u1, v1, v2) || dot_online_in(u2, v1, v2) || dot_online_in(v1, u1, u2) || dot_online_in(v2, u1, u2); } int main() { scanf("%d", &T); while (T--) { long long ans = 0; scanf("%d", &n); tot = 0; memset(flag, 0, sizeof flag); for (int i = 1; i <= n; i++) { scanf("%lf%lf", &p[tot].x, &p[tot].y); p[tot].id = i; tot++; scanf("%lf%lf", &p[tot].x, &p[tot].y); p[tot].id = i; tot++; } sort(p, p + tot, cmp); for (int i = 0; i < tot; i++) { if (!flag[p[i].id]) { flag[p[i].id] = 1; line[p[i].id].a = i; } else { line[p[i].id].b = i; p[i].id = -p[i].id; } } for (int i = 0; i < tot; i++) { if (p[i].id>0) { int j; for (j = i + 1; p[j].id != -p[i].id; j++) { if (p[j].id > 0) { if (intersect_in(p[line[p[i].id].a], p[line[p[i].id].b], p[line[p[j].id].a], p[line[p[j].id].b])) ans++; } } for (;; j++) { if (j+1<tot&&p[j].x == p[j + 1].x&&p[j].y == p[j + 1].y) { if (p[j+1].id>0) if (intersect_in(p[line[p[i].id].a], p[line[p[i].id].b], p[line[p[j+1].id].a], p[line[p[j+1].id].b])) ans++; } else break; } } } printf("%lld\n", ans); } return 0; }
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