Nested Segments(离散+树状数组)

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 ( - 109 ≤ li < ri ≤ 109) — 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.

Examples
inputCopy

4
1 8
2 3
4 7
5 6

outputCopy

3
0
1
0

inputCopy

3
3 4
1 5
2 6

outputCopy

0
1
1

#include 
#define ios ios::sync_with_stdio(false);cin.tie(0);
#define endl "\n"
const int N  = 1e7 + 5;
using namespace std;
int ans[N], b[N], re[N], n, m, k;
int lowbit(int x){
    return x & -x;
}
int update(int x, int m){
    for(int i = x; i <= k; i += lowbit(i)){
        ans[i] += m;
    }
}
int query(int x){
    int sum = 0;
    for(int i = x; i > 0; i-= lowbit(i)){
        sum += ans[i];
    }
    return sum;
}
int ls(int x){
    x = lower_bound(b + 1, b + 1 + m, x) - b;
    return x;
}
struct node{
    int l, r, pos;
    bool operator < (const node &x) const{//结构体内嵌比较函数
        if(l == x.l)
            return r > x.r;
        return l < x.l;
    }
}a[N];
int main(){
   int n;
   scanf("%d", &n);
   for(int i = 1; i <= n; i++){
      scanf("%d %d", &a[i].l, &a[i].r);
      a[i].pos = i;
      b[++k] = a[i].l;
      b[++k] = a[i].r;
   }
   sort(a + 1, a + 1 + n);
   sort(b + 1, b + 1 + k);
   m = unique(b + 1, b + 1 + k) - (b + 1);
  // ls()  离散, 求(左)右端点相对位置
  for(int i = 1; i <= n; i++)
 　   update(ls(a[i].r), 1);
  for(int i = 1; i <= n; i++){
   　 update(ls(a[i].r), -1);//delete
  　  re[a[i].pos] = query(ls(a[i].r));
  }
  for(int i = 1; i <= n; i++)
        printf("%d\n", re[i]);

}

离散模板：

for(int i = 1;i <= n; i++){
     cin>>a[i];
     b[i]=a[i];//创建两个数组
   }
   sort(b + 1, b + n + 1);//排序
   int m = unique(b + 1, b + n + 1) - b - 1;//去重
   for(int i = 1; i <= n; i++)
      a[i]=lower_bound(b + 1, b + m + 1, a[i]) - b;//减去排序后数字对应的位置, 可求其相对大小