HUST 1376 Random intersection

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
", ans);
    }
    return 0;
}