题意：

$给定N\le 2.5\times 10^4个矩形，保证没有重合，且任何边相切或者顶点重合的是不好的$
$问最后还有几个好的$

分析：

$- - 一眼扫描线了，关键是y轴怎么搞$
$蓝儿扫描线的时候自己太naive复杂度爆炸$
$事实上把矩形看作4个事件，y方向各自1个出入$
$然后我们发现相切的情况实际上就是前缀和\ge 2$
$嘛，注意顶点的情况，这个时候要先算进入事件$
$时间复杂度O(nlogn)$

$代码：$

//
//  Created by TaoSama on 2016-08-24
//  Copyright (c) 2016 TaoSama. All rights reserved.
//
#pragma comment(linker, "/STACK:102400000,102400000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>

using namespace std;
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;

int n;
int bx[N], by[N], ux[N], uy[N];

struct Line {
    int x, y, id;
    Line() {}
    Line(int x, int y, int id): x(x), y(y), id(id) {}
    bool operator<(const Line& l) const {
        if(x == l.x) {
            if(y == l.y) return id < l.id; //+ first
            return y < l.y;
        }
        return x < l.x;
    }
};

void solve(vector<int>& bad, int* bx, int* ux, int* by, int* uy) {
    vector<Line> events;
    for(int i = 1; i <= n; ++i) {
        events.push_back(Line(bx[i], by[i], -i));
        events.push_back(Line(bx[i], uy[i], i));
        events.push_back(Line(ux[i], by[i], -i));
        events.push_back(Line(ux[i], uy[i], i));
    }
    sort(events.begin(), events.end());

    int sum = 1;
    for(int i = 1; i < events.size(); ++i) {
        sum += events[i].id < 0 ? 1 : -1;
        int preID = abs(events[i - 1].id), curID = abs(events[i].id);
        if(sum >= 2) bad[preID] = bad[curID] = true;
    }
}

int main() {
#ifdef LOCAL
    freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
    ios_base::sync_with_stdio(0);

    while(scanf("%d", &n) == 1) {
        for(int i = 1; i <= n; ++i)
            scanf("%d%d%d%d", bx + i, by + i, ux + i, uy + i);

        vector<int> bad(n + 1, false);
        solve(bad, bx, ux, by, uy);
        solve(bad, by, uy, bx, ux);

        int ans = n - count(bad.begin(), bad.end(), true);
        printf("%d\n", ans);
    }

    return 0;
}