Educational Codeforces Round 10 D. Nested Segments 离线树状数组 离散化

D. Nested Segments

题目连接：

http://www.codeforces.com/contest/652/problem/D

Description

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.

Sample Input

4

1 8

2 3

4 7

5 6

Sample Output

3

0

1

0

Hint

题意

给你n个线段，让你输出每个线段完全包含多少个其他的线段

保证任意两个线段的端点不重合

题解：

先离散化一波

然后离线树状数组维护一波就好了

for循环暴力扫左端点，然后树状数组去查询小于右端点的线段有多少个，就完了

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 4e5+7;
pair<pair<int,int>,int>p[maxn];
int ans[maxn];
vector<int> Q;
map<int,int> H;
int a[maxn];
int lowbit(int x){return x&(-x);}
void update(int x,int v)
{
for(int i=x;i<maxn;i+=lowbit(i))
a[i]+=v;
}
int get(int x)
{
int tot = 0;
for(int i=x;i;i-=lowbit(i))
tot+=a[i];
return tot;
}
int main()
{
int n;
scanf("%d",&n);
for(int i=1;i<=n;i++)
{
scanf("%d%d",&p[i].first.first,&p[i].first.second);
Q.push_back(p[i].first.first),Q.push_back(p[i].first.second);
p[i].second=i;
}
sort(Q.begin(),Q.end());
Q.erase(unique(Q.begin(),Q.end()),Q.end());
for(int i=0;i<Q.size();i++)
H[Q[i]]=i+1;
for(int i=1;i<=n;i++)
{
p[i].first.first=H[p[i].first.first];
p[i].first.second=H[p[i].first.second];
update(p[i].first.second,1);
}
sort(p+1,p+1+n);
for(int i=1,j=1;i<maxn;i++)
{
while(j<=n&&p[j].first.first==i)
{
ans[p[j].second]=get(p[j].first.second);
update(p[j].first.second,-1);
j++;
}
if(j==n+1)break;
}
for(int i=1;i<=n;i++)
printf("%d\n",ans[i]-1);
}