codeforces 652D - Nested Segments

Posted 2020-08-16 Gealo

You are given n segments on a line. There are no ends of some segments that coincide. For each segment find the number of segments it contains.

 

Input

The first line contains a single integer n (1?≤?n?≤?2·105) — the number of segments on a line.

Each of the next n lines contains two integers li and ri (?-?10e9?≤?li?<?ri?≤?10e9) — the coordinates of the left and the right ends of the i-th segment. It is guaranteed that there are no ends of some segments that coincide.

 

Output

Print n lines. The j-th of them should contain the only integer aj — the number of segments contained in the j-th segment.

 

题意：给你n个区间问你每个区间包含了几个小区间，这些区间没有相交。

 求区间个数与线段树求逆序对方法差不多，先按右区间排序，再利用求逆序数的思想。还有由于区间数比较大但区间的个数少，要离散化一下。

 1 #include <iostream>
 2 #include <cstring>
 3 #include <algorithm>
 4 #include <map>
 5 using namespace std;
 6 const int M = 2e5 + 10;
 7 struct ss {
 8     int x , y , num;
 9 }s[M];
10 int m[M << 1] , ans[M << 1];
11 bool cmp(ss a , ss b) {
12     return a.y < b.y;
13 }
14 bool cmp1(ss a , ss b) {
15     return a.num < b.num;
16 }
17 struct TnT {
18     int l , r , sum;
19 }T1[M << 4] , T2[M << 4];
20 void build(TnT T[] , int l , int r , int p) {
21     int mid = (l + r) >> 1;
22     T[p].l = l , T[p].r = r , T[p].sum = 0;
23     if(T[p].l == T[p].r) {
24         return ;
25     }
26     build(T , l , mid , p << 1);
27     build(T , mid + 1 , r , (p << 1) | 1);
28 }
29 void updata(TnT T[] , int pos , int p) {
30     int mid =(T[p].l + T[p].r) >> 1;
31     if(T[p].l == T[p].r && T[p].l == pos) {
32         T[p].sum++;
33         return ;
34     }
35     if(mid >= pos) {
36         updata(T , pos , p << 1);
37     }
38     else {
39         updata(T , pos , (p << 1) | 1);
40     }
41     T[p].sum = T[p << 1].sum + T[(p << 1) | 1].sum;
42 }
43 int query(TnT T[] , int l , int r , int p) {
44     int mid = (T[p].l + T[p].r) >> 1;
45     if(T[p].l == l && T[p].r == r) {
46         return T[p].sum;
47     }
48     if(mid < l) {
49         return query(T , l , r , (p << 1) | 1);
50     }
51     else if(mid >= r) {
52         return query(T , l , r , p << 1);
53     }
54     else {
55         return query(T , l , mid , p << 1) + query(T , mid + 1 , r , (p << 1) | 1);
56     }
57 }
58 int main()
59 {
60     int n;
61     while(cin >> n) {
62         int MAX = 0;
63         map<int , int> mp;
64         int cnt = 0;
65         memset(s , 0 , sizeof(s));
66         for(int i = 0 ; i < n ; i++) {
67             cin >> s[i].x >> s[i].y;
68             m[++cnt] = s[i].x;
69             m[++cnt] = s[i].y;
70             s[i].num = i;
71             MAX = max(MAX , s[i].y);
72         }
73         sort(m + 1 , m + cnt + 1);
74         for(int i = 1 ; i <= cnt ; i++) {
75             mp[m[i]] = i;
76         }
77         build(T1 , 1 , cnt , 1);
78         build(T2 , 1 , cnt , 1);
79         sort(s , s + n , cmp);
80         for(int i = 0 ; i < n ; i++) {
81             updata(T1 , mp[s[i].y] , 1);
82             updata(T2 , mp[s[i].x] , 1);
83             ans[s[i].num] = min(query(T1 , 1 , mp[s[i].y] - 1 , 1) , query(T2 , mp[s[i].x] + 1 , cnt , 1));
84         }
85         for(int i = 0 ; i < n ; i++) {
86             cout << ans[i] << endl;
87         }
88     }
89     return 0;
90 }