Description
Input
Output
Sample Input
6 3 2 0 3 1 1 3 4 2 0 4 5 5
Sample Output
2
Data Constraint
Hint
贪心，把旅行者和出口的x坐标降序排序。
然后从前往后扫，如果是出口，就把y坐标插进set里，如果是旅行者，就查询set里是否有大于等于他纵坐标的值，有的话就答案加一，把这个值从set里删掉。
注意用multiset，并且不能删除一个空的迭代器。
#include <iostream> #include <cstdio> #include <algorithm> #include <cstring> #include <set> using namespace std; #define reg register inline char gc() {static const int BS = 1 << 22;static unsigned char buf[BS], *st, *ed;if (st == ed) ed = buf + fread(st = buf, 1, BS, stdin);return st == ed ? EOF : *st++; } #define gc getchar inline int read() {int res = 0;char ch=gc();bool fu=0;while(!isdigit(ch))fu|=(ch=='-'),ch=gc();while(isdigit(ch))res=(res<<3)+(res<<1)+(ch^48),ch=gc();return fu?-res:res; }int TT, n; struct date {int x, y, typ;bool operator < (const date &a) const {if (x == a.x) return y < a.y;return x > a.x;} }da[200005]; int cnt; int ans; multiset <int> s;int main() { // freopen("guide.in", "r", stdin); // freopen("guide.out", "w", stdout);TT = read(), n = read();for (reg int i = 1 ; i <= n ; i ++){int x = read(), y = read();da[++cnt] = (date){x, y, 1};}for (reg int i = 1 ; i <= n ; i ++){int x = read(), y = read();da[++cnt] = (date){x, y, 2};}sort(da + 1, da + 1 + cnt);for (reg int i = 1 ; i <= cnt ; i ++){if (da[i].typ == 2) s.insert(da[i].y);else if (!s.empty()) {set <int> :: iterator it = s.lower_bound(da[i].y);if (*it >= da[i].y) s.erase(*it), ans++;}}cout << ans << endl;return 0; }
