codeforces 1284D. New Year and Conference (segment tree)

Link: https: //codeforces.com/problemset/problem/1284/D

The meaning of problems: there are n lectures, there are two sites a and b, if the site is open talks in a need to occupy [SAI, EAI], lectures need to occupy the space in the b [sbi, ebi] this time, if the open two sessions, if a site does not open in the conflict, and b venues open conflict, claimed it as sensitive, empathy and a and b and vice versa, if the two are ab conflict is not a sensitive, seek first n lectures given any pairwise whether to open sensitive.

Ideas: violence enumeration is O (n ^ 2) complexity, will time-out, where you can use ST segment tree or table to do, achieve a complexity of nlogn. For any x lectures, seminars and find all sites x lecture in a conflict, then it is determined whether the conflict in site B, if not directly output "NO"

Interval after the first all lectures sa and ea ascending, such as a time slice lecture i is [x, y], in its time slice and i lecture a space conflict with the method of two points can be enumerated a discretization then we sb maximum and minimum maintenance b eb site with tree line, every check out this section

Sb maximum and minimum values of eb, at this time if the maximum sb> y or the minimum value eb <x, then the presence of these performances must be b i and performing space does not conflict, then direct output "NO", the All lectures will check it again, and then check the order ab exchange to the test only in a certain field of conflict in the b field of conflict situations, but not in the b field conflict is not a conflict in a venue to consider the situation.

The overall time complexity nlogn.

AC Code:

 1 #include<iostream>
 2 #include<vector>
 3 #include<cstdlib>
 4 #include<cstdio>
 5 #include<algorithm>
 6 #include<cmath>
 7 #include<cstring>
 8 #include<queue>
 9 #include<map>
10 using namespace std;
11 typedef long long ll;
12 const int maxn = 1e5+200;
13 struct node{
14     int sa,ea,sb,eb;
15     node(){};
16     node(int a,int b,int c,int d){
17         sa = a,ea = b,sb = c,eb = d;
18     }
19     bool operator<(node cur)const{
20         if(sa == cur.sa ) return ea<cur.ea ;
21         return sa < cur.sa ;
22     }
23 }point[maxn];
24 int segt_max[4*maxn],segt_min[4*maxn];
25 void build(intL, int R & lt, int K) { // build two segment tree maximum and minimum maintenance interval 
26 is      IF (L == R & lt) {
 27          segt_max [K] = Point [L] .sb;
 28          segt_min [K] = Point [L] .eb;
 29          return ; 
 30      }
 31 is      int MID = (L + R & lt) / 2 ;
 32      Build (L, MID, 2 * K);
 33 is      Build (MID + . 1 , R & lt, 2 * K + . 1 );
 34 is      segt_max [K] = max (segt_max [ 2 * K], segt_max [ 2* K + . 1 ]);
 35      segt_min [K] = min (segt_min [ 2 * K], segt_min [ 2 * K + . 1 ]);
 36  }
 37 [  int queryMin ( int L, int R & lt, int Al, int Ar, int K ) { // query minimum interval 
38 is      IF (L> = R & lt && Al <= Ar) return segt_min [K];
 39      int MID = (L + R & lt) / 2 ;
 40      IF (Ar <= MID) return queryMin ( L, MID, Al, Ar, 2 * K);
 41 is     else if(al>mid){
42         return queryMin(mid+1,r,al,ar,2*k+1);
43     }
44     else {
45         return min(queryMin(l,mid,al,mid,2*k),queryMin(mid+1,r,mid+1,ar,2*k+1));
46     }
47 } 
48 int queryMax(int l,int r,int al,int ar,int k){//查询区间最大值 
49     if(l >= al && r <= ar) return segt_max[k];
50     int mid = (l+r)/2;
51     if(ar<=mid) return queryMax(l,mid,al,ar,2*k);
52     else if(al>mid){
53         return queryMax(mid+1,r,al,ar,2*k+1);
54     }
55     else {
56         return max(queryMax(l,mid,al,mid,2*k),queryMax(mid+1,r,mid+1,ar,2*k+1));
57     }
58 } 
59 bool check(int n){
60     sort(point+1,point+1+n);
61     build(1,n,1);
62     for(int i = 1;i<=n;i++){
63         int pos = lower_bound(point+1,point+1+n,node(point[i].ea ,1e9+5000,0,0))-point-1;
 64          // Find the i-th set of speeches conflict 
65          IF (i + 1 > POS) the Continue ; // no conflict skip 
 66          // the Check to see if all the other performance venues are conflict 
67          IF (queryMin ( 1 , n-, I + . 1 , POS, . 1 ) <Point [I] .sb queryMax || ( . 1 , n-, I + . 1 , POS, . 1 )> Point [I] .eb) {
 68              return  to false ;
 69          }
 70      }
 71 is      return  to true ;
 72  }
 73 is  int main()
74 {
75     int n;
76     cin>>n;
77     int f = 0;
78     for(int i = 1;i<=n;i++){
79         cin>>point[i].sa>>point[i].ea>>point[i].sb>>point[i].eb;    
80     }
81     if(check(n)) f++;
82     for(int i = 1;i<=n;i++){
83         swap(point[i].sa,point[i].sb);
84         swap(point[i].ea,point[i].eb);
85     }
86     if(check(n)) f++;
87     if(f == 2) cout<<"YES";
88     else cout<<"NO"; 
89     return 0;
90 }