[JZOJ5866]【NOIP2018模拟9.13】指引

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;
}