ZOJ 3157 Weapon

 

题目传送门

题意：就是CF round# 329 B 的升级版，要求出相交点的个数

分析：逆序数用树状数组维护，求出非逆序数，然后所有情况(n * (n - 1)) / 2减之就是逆序数个数。

 

#include <bits/stdc++.h>
using namespace std;

const int N = 1e4 + 10;
const double EPS = 1e-8;

double x[N][2], y[N][2];
pair<double, double> a[N];
double A[N];

struct BIT	{
	int c[N], sz;
	void init(int n)	{
		sz = n;
		memset (c, 0, sizeof (c));
	}
	void updata(int i, int x)	{
		while (i <= sz)	{
			c[i] += x;	i += i & (-i);
		}
	}
	int query(int i)	{
		int ret = 0;
		while (i)	{
			ret += c[i];	i -= i & (-i);
		}
		return ret;
	}
}bit;

int dcmp(double x)	{
	if (fabs (x) < EPS)	return 0;
	else	return x < 0 ? -1 : 1;
}

double cross(int i, double X)	{
//	if (dcmp (x[i][0] - x[i][1]) == 0)	return 0;
    double k = y[i][1] - y[i][0];
    k /= (x[i][1] - x[i][0]);
    double b = -k * x[i][0] + y[i][0];
    return k * X + b;
}

int main(void)	{
	int n;
	while (scanf ("%d", &n) == 1)	{
		for (int i=0; i<n; ++i)	{
			scanf ("%lf%lf%lf%lf", &x[i][0], &y[i][0], &x[i][1], &y[i][1]);
		}
		double L, R;	scanf ("%lf%lf", &L, &R);
		L += EPS;	R -= EPS;
		for (int i=0; i<n; ++i)	{
			a[i].first = cross (i, L);
			a[i].second = cross (i, R);
			cout << a[i].first << " " << a[i].second << endl;
		}
		sort (a, a+n);
		for (int i=0; i<n; ++i)	A[i] = a[i].second;
		sort (A, A+n);
		int tot = unique (A, A+n) - A;
		bit.init (n);
		int ans = 0;
		for (int i=0; i<n; ++i)	{
			int pos = lower_bound (A, A+tot, a[i].second) - A + 1;
			ans += bit.query (pos);
			bit.updata (pos, 1);
		}
		printf ("%d\n", n * (n - 1) / 2 - ans);
	}

	return 0;
}