codeforces Hello 2020 D. New Year and Conference

（还没理解）

题意：

Filled with optimism, Hyunuk will host a conference about how great this new year will be!

The conference will have $n$ lectures. Hyunuk has two candidate venues $a$ and $b$ . For each of the $n$ lectures, the speaker specified two time intervals $\text{[math]}$ and $\text{[math]}$ . If the conference is situated in venue $a$ , the lecture will be held from $sai$ to $eai$ , and if the conference is situated in venue $b$ , the lecture will be held from $sbi$ to $ebi$ . Hyunuk will choose one of these venues and all lectures will be held at that venue.

Two lectures are said to overlap if they share any point in time in common. Formally, a lecture held in interval $\text{[math]}$ overlaps with a lecture held in interval $\text{[math]}$ if and only if $\text{[math]}$ .

We say that a participant can attend a subset $s$ of the lectures if the lectures in $s$ do not pairwise overlap ( $i.e.$ no two lectures overlap). Note that the possibility of attending may depend on whether Hyunuk selected venue $a$ or venue $b$ to hold the conference.

A subset of lectures $s$ is said to be venue-sensitive if, for one of the venues, the participant can attend $s$ , but for the other venue, the participant cannot attend $s$ .

A venue-sensitive set is problematic for a participant who is interested in attending the lectures in $s$ because the participant cannot be sure whether the lecture times will overlap. Hyunuk will be happy if and only if there are no venue-sensitive sets. Determine whether Hyunuk will be happy.

Input

The first line contains an integer $n$ ( $\text{[math]}$ ), the number of lectures held in the conference.

Each of the next $n$ lines contains four integers $\text{[math]}$

Output

Print "YES" if Hyunuk will be happy. Print "NO" otherwise.

You can print each letter in any case (upper or lower).

Examples

input

2
1 2 3 6
3 4 7 8

output

YES

input

output

NO

input

output

YES

Note

In second example, lecture set $\text{[math]}$ is venue-sensitive. Because participant can't attend this lectures in venue $a$ , but can attend in venue $b$ .

In first and third example, venue-sensitive set does not exist.

题意：

有 $2n$ 个区间，分别为 $\text{[math]}$ ，每两个区间为一对，共 $n$ 对区间。一对中的两个区间绑定在一起，从 $n$ 对中选出一个子集。存在一种情况是：子集中某个系列（ $a$ 或 $b$ ）区间有区间交，而另外一个系列的区间没有区间交。如果这个情况发生在某个子集中，则输出NO,否则输出YES

思路：

先考虑 $a$ 系列区间交的所有可能，枚举交点，将包含该交点的所有 $a$ 区间都放到数轴上，同时把对应的 $b$ 区间也放在数轴上，那么这些 $b$ 区间中两两之间也必须在某点交。

#include <bits/stdc++.h>
using namespace std;
#define ll long long
const int maxn = 4e5 + 10;
const int inf = 0x3f3f3f3f;
int n;
struct SegTree{
    int l, r;
    int mx, lazy;
} t[maxn * 4];
struct node{
    int l, r, id;
} a[maxn], b[maxn];
vector<int> l[maxn], r[maxn], alls;

void build(int p, int l, int r){
    t[p].l = l;
    t[p].r = r;
    if(l == r){
        t[p].mx = t[p].lazy = 0;
        return;
    }
    int mid = (l + r) >> 1;
    build(p << 1, l, mid);
    build(p << 1 | 1, mid + 1, r);
    t[p].mx = t[p].lazy = 0;
}

void pushdown(int p){
    if(t[p].lazy){
        t[p << 1].mx += t[p].lazy;
        t[p << 1 | 1].mx += t[p].lazy;
        t[p << 1].lazy += t[p].lazy;
        t[p << 1 | 1].lazy += t[p].lazy;
        t[p].lazy = 0;
    }
}

int ask(int p, int l, int r){
    if(t[p].l >= l && t[p].r <= r){
        return t[p].mx;
    }
    pushdown(p);
    int mid = (t[p].l + t[p].r) >> 1;
    int res = 0;
    if(mid >= l)
        res = max(res, ask(p << 1, l, r));
    if(mid<r)
        res = max(res, ask(p << 1 | 1, l, r));
    return res;
}

void add(int p, int l, int r, int val){
    if(t[p].l >= l && t[p].r <= r){
        t[p].mx += val;
        t[p].lazy += val;
        return;
    }
    pushdown(p);
    int mid = (t[p].l + t[p].r) >> 1;
    if(mid >= l)
        add(p << 1, l, r, val);
    if(mid<r)
        add(p << 1 | 1, l, r, val);
    t[p].mx = max(t[p << 1].mx, t[p << 1 | 1].mx);
}

bool check(node *a, node *b){
    int len = alls.size();
    build(1, 1, len);
    for (int i = 1; i <= len; i++){
        l[i].clear();
        r[i].clear();
    }
    for (int i = 1; i <= n; i++){
        l[a[i].l].push_back(i);
        r[a[i].r].push_back(i);
    }

    int sz = 0;
    for (int i = 1; i <= len; i++){
        for (int j = 0; j < l[i].size(); j++){
            int id = l[i][j];

            int L = b[id].l, R = b[id].r;

            int res = ask(1, L, R);
            if(res != sz)
                return false;
            add(1, L, R, 1);
            sz++;
        }
        for (int j = 0; j < r[i].size(); j++){
            int id = r[i][j];
            int L = b[id].l, R = b[id].r;
            add(1, L, R, -1);
            sz--;
        }
    }
    return true;
}

int getId(int x){
    return lower_bound(alls.begin(), alls.end(), x) - alls.begin() + 1;
}
int main(){
    ios_base::sync_with_stdio(0);
    cin >> n;
    for (int i = 1; i <= n; i++){
        cin >> a[i].l >> a[i].r >> b[i].l >> b[i].r;
        alls.push_back(a[i].l);
        alls.push_back(a[i].r);
        alls.push_back(b[i].l);
        alls.push_back(b[i].r);
        a[i].id = i;
        b[i].id = i;
    }
    sort(alls.begin(), alls.end());
    alls.erase(unique(alls.begin(), alls.end()), alls.end());
    for (int i = 1; i<=n; i++){
        a[i].l = getId(a[i].l);
        a[i].r = getId(a[i].r);
        b[i].l = getId(b[i].l);
        b[i].r = getId(b[i].r);
    }
    int c1 = check(a, b);
    int c2 = check(b, a);
    if(c1 && c2)
        cout << "YES" << endl;
    else
        cout << "NO" << endl;
    return 0;
}