题干：
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.
题目大意：
给定n个区间线段，要求输出n个数，其中第i个数代表：被i个线段覆盖的点的个数。
解题报告：
离散化差分，跟2019icpc上海网络赛那题一样的套路。
AC代码：
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define FF first
#define SS second
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
typedef pair<ll,ll> PLL;
const int MAX = 2e5 + 5;
map<ll,int> mp;
int n;
ll l,r;
ll ans[MAX];//下标表示覆盖次数
int main()
{cin>>n;for(int i = 1; i<=n; i++) cin>>l>>r,mp[l]++,mp[r+1]--;ll pre = -1,sum=0;for(auto it = mp.begin(); it != mp.end(); ++it) {ans[sum] += it->FF - pre;sum += it->SS; pre = it->FF; }for(int i = 1; i<=n; i++) {printf("%lld%c",ans[i],i == n ?'\n' :' ');}return 0 ;
}
