Nested Ranges Count

题目描述

Given n ranges, your task is to count for each range how many other ranges it contains and how many other ranges contain it.

Range $a,b$ contains range $c,d$ if a ≤ c and d ≤ b.

输入

The first input line has an integer n(1 ≤ n ≤ 2*1e5): the number of ranges.

After this, there are n lines that describe the ranges. Each line has two integers x and y(1 ≤ x < y ≤ 1e9): the range is $x,y$ .

You may assume that no range appears more than once in the input.

输出

First print a line that describes for each range (in the input order) how many other ranges it contains.

Then print a line that describes for each range (in the input order) how many other ranges contain it.

样例输入

4

1 6

2 4

4 8

3 6

样例输出

2 0 0 0

0 1 0 1

题目大意：n个区间，判断每个区间包含了多少个区间，以及每个区间被多少个区间包含

思路：一个区间 $l,r$ 包含其他区间，意味着它的 l 要更小，r 要更大，按照 l 升序，r 降序进行排序，排序后，每个区间右边的，都要比它的 l 小，只需要找它右边有多少个区间的 r 比它大即可，我们可以考虑逆序遍历，先处理右边的，将每个区间的 r 记下来，最后输出值在1到a $i$ .r之间的数量，这个过程可以用树状数组维护。但由于 r 达到了1e9，个数只有2e5，离散化处理即可。

代码

cpp 复制代码

#include <bits/stdc++.h>
using namespace std;
using ll=long long;
const int N=200010;
int tr[N];
int n,m;
int containing[N],contained[N];
struct Range{
    int l,r,idx;
}a[N];
int lowbit(int x){
    return x&-x;
}
void update(int k,int x){
    for (int i=k;i<=n;i+=lowbit(i)){
        tr[i]+=x;
    }
}
int query(int x){
    int res=0;
    for(int i=x;i>=1;i-=lowbit(i)){
        res+=tr[i];
    }
    return res;
}
int main() {
    ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    cin>>n;
    vector<int>alls;
    for (int i=1;i<=n;i++){
        cin>>a[i].l>>a[i].r;
        a[i].idx=i;
        alls.push_back(a[i].r);
    }
    sort(alls.begin(),alls.end());
    alls.erase(unique(alls.begin(),alls.end()),alls.end());
    m=alls.size();
    sort(a+1,a+n+1,[](const struct Range &x,const struct Range &y){
        if(x.l==y.l)
        return x.r>y.r;
        return x.l<y.l;
    });
    for (int i=n;i>=1;i--){
        int t=lower_bound(alls.begin(),alls.end(),a[i].r)-alls.begin()+1;
        containing[a[i].idx]=query(t);
        update(t,1);
    }
    memset(tr,0,sizeof(tr));
    for (int i=1;i<=n;i++){
        int t=lower_bound(alls.begin(),alls.end(),a[i].r)-alls.begin()+1;
        contained[a[i].idx]=query(m)-query(t-1);
        update(t,1);
    }
    for (int i=1;i<=n;i++)
    cout<<containing[i]<<' ';
    cout<<"\n";
    for (int i=1;i<=n;i++)
    cout<<contained[i]<<' ';
    return 0;
}