C. Covered Points Count
time limit per test
3 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
You are given nn 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∈[1..n]k∈[1..n], calculate the number of points with integer coordinates such that the number of segments that cover these points equals kk. A segment with endpoints lili and riri covers point xx if and only if li≤x≤rili≤x≤ri.
Input
The first line of the input contains one integer nn (1≤n≤2⋅1051≤n≤2⋅105) — the number of segments.
The next nn lines contain segments. The ii-th line contains a pair of integers li,rili,ri (0≤li≤ri≤10180≤li≤ri≤1018) — the endpoints of the ii-th segment.
Output
Print nn space separated integers cnt1,cnt2,…,cntncnt1,cnt2,…,cntn, where cnticnti is equal to the number of points such that the number of segments that cover these points equals to ii.
Examples
input
Copy
3 0 3 1 3 3 8
output
Copy
6 2 1
input
Copy
3 1 3 2 4 5 7
output
Copy
5 2 0
Note
The picture describing the first example:
Points with coordinates [0,4,5,6,7,8][0,4,5,6,7,8] are covered by one segment, points [1,2][1,2] are covered by two segments and point [3][3] is covered by three segments.
The picture describing the second example:
Points [1,4,5,6,7][1,4,5,6,7] are covered by one segment, points [2,3][2,3] are covered by two segments and there are no points covered by three segments.
扫描到的当前边界减去上一次的边界，得到标记和对应的点数
#include <iostream>
#include <cstring>
#include <string>
#include <map>
#include <queue>
#include <stack>
#include <deque>
using namespace std;
#define ll long long
#define N 200005
map<ll,int > mapp;
map<ll,int > ::iterator it;
ll cnt[N],n,l,r;
int main()
{
cin>>n;
for(int i=0;i<n;i++)
{
cin>>l>>r;
mapp[l]++;//起点， 表示 增加了一条覆盖的线段。
mapp[r+1]--;//因为是闭区间，所以应该往后移一位，表示少了一条覆盖的线段。
}
ll st=0;//当前区间的起点
ll ans=0;
for(it=mapp.begin();it!=mapp.end();it++)
{
//cout<<it->first<<" "<<it->second<<endl;
//cout<<"** st:"<<st<<endl;
if(it==mapp.begin())
{
st=it->first;//起点 map<ll,int > mapp 中的 ll
ans+=it->second;//map<ll ,int > mapp 中的 int 表示 此点处线段覆盖数
continue;
}
// cout<<"*** it->first-st:"<<it->first - st<<" "<<endl;
cnt[ans]+=it->first - st;//上一个起点 减去 下一个起点 中间的 线段数
//cout<<"** cnt[ans]:"<<ans<<" "<<cnt[ans]<<endl;
st=it->first;
ans+=it->second;
}
for(int i=1;i<=n;i++)
{
cout<<cnt[i]<<" ";
}return 0;
}