本文共 2183 字，大约阅读时间需要 7 分钟。

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

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:

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.

要是之前做过这道题，上海网络赛那道题就不会卡那么久了。。 和那道题一样的思路，离散化之后差分就可以了。

#include
   
    #define ll long longusing namespace std;const int maxx=2e5+100;struct node{
   	ll x,y;}p[maxx];ll a[maxx<<1];int b[maxx<<1];ll c[maxx];int n;int main(){
   	scanf("%d",&n);	int cnt=0,x,y;	for(int i=1;i<=n;i++)	{
   		scanf("%lld%lld",&p[i].x,&p[i].y);		a[++cnt]=p[i].x;		a[++cnt]=p[i].y+1;	}	sort(a+1,a+1+cnt);	int len=unique(a+1,a+1+cnt)-a-1;	for(int i=1;i<=n;i++)	{
   		x=lower_bound(a+1,a+1+len,p[i].x)-a;		y=lower_bound(a+1,a+1+len,p[i].y+1)-a;		b[x]++;b[y]--;	}	for(int i=1;i<=len;i++)	{
   		b[i]+=b[i-1];		c[b[i]]+=a[i+1]-a[i];//c数组注意用long long，这里代表着从a[i]开始到a[i+1]都是被染色了b[i]次。	}	for(int i=1;i<=n;i++) printf("%lld%c",c[i],i==n?'\n':' ');}
   

努力加油a啊，(^o)/~

转载地址：http://yytvi.baihongyu.com/