HUST 1376 Random intersection

Description

There are N segments on the plane. You are to find out how many pairs of segments intersect with each other. A pair of segments intersect with each other if and only if they have at least one common point.

Input

the first line contains a single integer T -- the number of the testcases then T testcases are followed each testcase contains an integer N( N <= 100000), the number of segments then N lines are followed, each line contains 4 floating point numbers: x1 y1 x2 y2 which means the coordinates of the two ends of a segment for more details, see the sample input (the input data is generated randomly, and the probability of a pair of segments intersecting is about 0.001)

Output

for each testcase, print 1 line with the number of the pairs of segments intersect with each other.

Sample Input

330.0000 0.0000 1.0000 2.00000.0000 1.0000 1.0000 0.00000.0000 2.0000 1.0000 1.000040.0000 0.0000 0.0000 -1.00000.0000 0.0000 1.0000 0.00000.0000 0.0000 0.0000 1.00000.0000 0.0000 -1.0000 0.000020.0000 0.0000 1.0000 1.00000.0000 0.0000 2.0000 2.0000

Sample Output

261

丧心病狂的题，能想到的只有暴力，然而单纯的暴力是不行的。

像线段树的扫描线一样做，把线段的端点标上号然后排序，只有在左右端点中间的点才可能和线段有交点。

然后暴力的判断就可以了，需要注意的是端点可能会有重合，需要特判一下。

#include<cstdio>#include<cstring>#include<cmath>#include<queue>#include<vector>#include<iostream>#include<algorithm>#include<bitset>#include<functional>using namespace std;typedef unsigned long long ull;typedef long long LL;const int maxn = 1e5 + 10;int T, n;struct point{double x, y;int id;void read(int z){ id = z; scanf("%lf%lf", &x, &y); }bool operator==(const point&a)const{ return x == a.x&&y == a.y; }bool operator<(const point&a)const{ return x == a.x ? y < a.y : x < a.x; }}L[maxn], R[maxn], a[maxn * 2];double mult(point a, point b, point c){return (a.x - c.x)*(b.y - c.y) - (b.x - c.x)*(a.y - c.y);}bool check(point aa, point bb, point cc, point dd){if (aa == cc&bb == dd) return true;if (max(aa.x, bb.x)<min(cc.x, dd.x)) return false;if (max(aa.y, bb.y)<min(cc.y, dd.y)) return false;if (max(cc.x, dd.x)<min(aa.x, bb.x)) return false;if (max(cc.y, dd.y)<min(aa.y, bb.y)) return false;if (mult(cc, bb, aa)*mult(bb, dd, aa)<0) return false;if (mult(aa, dd, cc)*mult(dd, bb, cc)<0) return false;return true;}int main(){cin >> T;while (T--){scanf("%d", &n);for (int i = 1; i <=n; i++){L[i].read(i);R[i].read(i);if (R[i] < L[i]) swap(L[i], R[i]);a[i * 2 - 1] = L[i];a[i * 2] = R[i];a[i * 2].id = -i;}sort(a + 1, a + n + n + 1);LL ans = 0;for (int i = 1; i <= n + n; i++){if (a[i].id > 0){int j;for (j = i + 1; -a[j].id != a[i].id; j++){if (a[j].id > 0 && check(L[a[i].id], R[a[i].id], L[a[j].id], R[a[j].id])) ans++;}for (j++; a[j] == a[j - 1]; j++)if (a[j].id > 0 && check(L[a[i].id], R[a[i].id], L[a[j].id], R[a[j].id])) ans++;}}printf("%lld\n", ans);}return 0;}