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。

 

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#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; }