codeforces Hello 2020 D. New Year and Conference

（還沒理解）

題意：

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

The conference will have lectures. Hyunuk has two candidate venues and . For each of the lectures, the speaker specified two time intervals and . If the conference is situated in venue , the lecture will be held from to , and if the conference is situated in venue , the lecture will be held from to . 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 overlaps with a lecture held in interval if and only if .

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

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

A venue-sensitive set is problematic for a participant who is interested in attending the lectures in 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 (), the number of lectures held in the conference.

Each of the next lines contains four integers

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

3
1 3 2 4
4 5 6 7
3 4 5 5

output

NO

input

6
1 5 2 9
2 4 5 8
3 6 7 11
7 10 12 16
8 11 13 17
9 12 14 18

output

YES

Note

In second example, lecture set is venue-sensitive. Because participant can't attend this lectures in venue , but can attend in venue .

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

題意：

有個區(qū)間，分別為，每兩個區(qū)間為一對，共對區(qū)間。一對中的兩個區(qū)間綁定在一起，從對中選出一個子集。存在一種情況是：子集中某個系列（或）區(qū)間有區(qū)間交，而另外一個系列的區(qū)間沒有區(qū)間交。如果這個情況發(fā)生在某個子集中，則輸出NO,否則輸出YES

思路：

先考慮系列區(qū)間交的所有可能，枚舉交點，將包含該交點的所有區(qū)間都放到數(shù)軸上，同時把對應(yīng)的區(qū)間也放在數(shù)軸上，那么這些區(qū)間中兩兩之間也必須在某點交。

#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;
}