C. Covered Points Count

最新推荐文章于 2021-03-25 11:38:25 发布

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最新推荐文章于 2021-03-25 11:38:25 发布

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本文介绍了一个经典的区间覆盖问题，给出了一种高效的算法解决方案。通过记录每个区间的左右端点，并对其进行特殊标记，最终统计出被不同数量的区间覆盖的整数坐标点的数量。

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You are given $n$ segments on a coordinate line; each endpoint of every segment has integer coordinates. Some segments can degenerate to points. Segments can intersect with each other, be nested in each other or even coincide.

Your task is the following: for every $k \in [1.. n]$ , calculate the number of points with integer coordinates such that the number of segments that cover these points equals $k$ . A segment with endpoints $l_{i}$ and $r_{i}$ covers point $x$ if and only if $l_{i} \leq x \leq r_{i}$ .

Input

The first line of the input contains one integer $n$ ( $1 \leq n \leq 2 \cdot 10^{5}$ ) — the number of segments.

The next $n$ lines contain segments. The $i$ -th line contains a pair of integers $l_{i}, r_{i}$ ( $0 \leq l_{i} \leq r_{i} \leq 10^{18}$ ) — the endpoints of the $i$ -th segment.

Output

Print $n$ space separated integers $c n t_{1}, c n t_{2}, \dots, c n t_{n}$ , where $c n t_{i}$ is equal to the number of points such that the number of segments that cover these points equals to $i$ .

Examples
inputCopy
3
0 3
1 3
3 8
outputCopy
6 2 1 
inputCopy
3
1 3
2 4
5 7
outputCopy
5 2 0 

Note

The picture describing the first example:

$\text{[math]}$

Points with coordinates $[0, 4, 5, 6, 7, 8]$ are covered by one segment, points $[1, 2]$ are covered by two segments and point $[3]$ is covered by three segments.

The picture describing the second example:

$\text{[math]}$

Points $[1, 4, 5, 6, 7]$ are covered by one segment, points $[2, 3]$

are covered by two segments and there are no points covered by three segments

求被i个区间覆盖的点的个数（1<=i<=n）.把每个点的左右端点记下来，左端点标记值+1,右端点后一位值-1。

怎么说还是看代码吧，想了很久没想出来怎么做

#include<bits/stdc++.h>
typedef long long ll;
using namespace std;
const int maxn = 500005;
pair<ll,ll>e[maxn];//存下了两个属性
ll f[maxn],n;
int main()
{
    ll l,r,cnt=1;
    scanf("%I64d",&n);
    ll ri=0;
    for(int i=1;i<=n;i++)
    {
        scanf("%I64d%I64d",&l,&r);
        e[++ri]={l,1};
        e[++ri]={r+1,-1};
    }
    sort(e+1,e+1+ri);
    for(int i=2;i<=ri;i++)
    {
        f[cnt]+=e[i].first-e[i-1].first;
        cnt+=e[i].second;//
    }
    printf("%I64d",f[1]);
    for(int i=2;i<=n;i++)
        printf(" %I64d",f[i]);
    printf("\n");
}