codeforces 652D - Nested Segments

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。

 

#include <iostream>
#include <cstring>
#include <algorithm>
#include <map>
using namespace std;
const int M = 2e5 + 10;
struct ss {
    int x , y , num;
}s[M];
int m[M << 1] , ans[M << 1];
bool cmp(ss a , ss b) {
    return a.y < b.y;
}
bool cmp1(ss a , ss b) {
    return a.num < b.num;
}
struct TnT {
    int l , r , sum;
}T1[M << 4] , T2[M << 4];
void build(TnT T[] , int l , int r , int p) {
    int mid = (l + r) >> 1;
    T[p].l = l , T[p].r = r , T[p].sum = 0;
    if(T[p].l == T[p].r) {
        return ;
    }
    build(T , l , mid , p << 1);
    build(T , mid + 1 , r , (p << 1) | 1);
}
void updata(TnT T[] , int pos , int p) {
    int mid =(T[p].l + T[p].r) >> 1;
    if(T[p].l == T[p].r && T[p].l == pos) {
        T[p].sum++;
        return ;
    }
    if(mid >= pos) {
        updata(T , pos , p << 1);
    }
    else {
        updata(T , pos , (p << 1) | 1);
    }
    T[p].sum = T[p << 1].sum + T[(p << 1) | 1].sum;
}
int query(TnT T[] , int l , int r , int p) {
    int mid = (T[p].l + T[p].r) >> 1;
    if(T[p].l == l && T[p].r == r) {
        return T[p].sum;
    }
    if(mid < l) {
        return query(T , l , r , (p << 1) | 1);
    }
    else if(mid >= r) {
        return query(T , l , r , p << 1);
    }
    else {
        return query(T , l , mid , p << 1) + query(T , mid + 1 , r , (p << 1) | 1);
    }
}
int main()
{
    int n;
    while(cin >> n) {
        int MAX = 0;
        map mp;
        int cnt = 0;
        memset(s , 0 , sizeof(s));
        for(int i = 0 ; i < n ; i++) {
            cin >> s[i].x >> s[i].y;
            m[++cnt] = s[i].x;
            m[++cnt] = s[i].y;
            s[i].num = i;
            MAX = max(MAX , s[i].y);
        }
        sort(m + 1 , m + cnt + 1);
        for(int i = 1 ; i <= cnt ; i++) {
            mp[m[i]] = i;
        }
        build(T1 , 1 , cnt , 1);
        build(T2 , 1 , cnt , 1);
        sort(s , s + n , cmp);
        for(int i = 0 ; i < n ; i++) {
            updata(T1 , mp[s[i].y] , 1);
            updata(T2 , mp[s[i].x] , 1);
            ans[s[i].num] = min(query(T1 , 1 , mp[s[i].y] - 1 , 1) , query(T2 , mp[s[i].x] + 1 , cnt , 1));
        }
        for(int i = 0 ; i < n ; i++) {
            cout << ans[i] << endl;
        }
    }
    return 0;
}