CodeForces 295E - Yaroslav and Points 题解：又是线段树！

Description

Yaroslav has n points that lie on the $Ox$ axis. The coordinate of the first point is $x_1$ , the coordinate of the second point is $x_2$ , ..., the coordinate of the n-th point is — $x_n$ . Now Yaroslav wants to execute $m$ queries, each of them is of one of the two following types:

Move the $p_j$ -th point from position $x_{p_j}$ to position . At that, it is guaranteed that after executing such query all coordinates of the points will be distinct. Count the sum of distances between all pairs of points that lie on the segment . In other words, you should count the sum of:

Help Yaroslav.

Link

time limit per test: 5 seconds
memory limit per test: 256 megabytes
input: standard input
output: standard output

Input

The first line contains integer $n$ — the number of points ( $1 ≤ n ≤ 10^5$ ). The second line contains distinct integers $x_1, x_2, \dots , x_n$ — the coordinates of points ().

The third line contains integer $m$ — the number of queries ( $1 ≤ m ≤ 10^5$ ). The next $m$ lines contain the queries. The $j$ -th line first contains integer () — the query type. If , then it is followed by two integers and (). If , then it is followed by two integers and ().

It is guaranteed that at any moment all the points have distinct coordinates.

Output

For each type 2 query print the answer on a single line. Print the answers in the order, in which the queries follow in the input.

Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams of the %I64d specifier.

Examples

Input

8
36 50 28 -75 40 -60 -95 -48
20
2 -61 29
1 5 -53
1 1 429
1 5 130
2 -101 -71
2 -69 53
1 1 404
1 5 518
2 -101 53
2 50 872
1 1 -207
2 -99 -40
1 7 -389
1 6 -171
1 2 464
1 7 -707
1 1 -730
1 1 560
2 635 644
1 7 -677

Output

Translation

题意：一维数轴上有 $n$ 个点，每个点有坐标 $x_i$ ，现在对这个数轴上的点有两种操作：

改变某个坐标 $x_i$ ，将其增加 $d$ （每次操作 d 不同）：
查询 $[L_i,R_i]$ 区间内的，也就是两两点对距离之和。

Analysis

又双叒叕是线段树。
可以写个单点修改区间查询的线段树。首先由于数据规模过大，需要离散化，把操作离线下来，把所有可能出现的值（包括初始、修改和查询）全都离散。现在我们可以对于这个一维数轴建一棵线段树，对于每个点如果有则值为 1，否则为 0。那么对于每个区间（线段树的每个节点），需要存储以下三个信息：

$sum_p$ ：这个区间中所有点坐标之和；
$tree_p$ ：这个区间中的 $\sum_{x_i>x_j} x_i-x_j$ ；

$tree_p=\sum_{x_i>x_j} x_i-x_j = \sum_{i=1}^n \sum_{j=1}^n x_i - \sum_{i=1}^n \sum_{j=1}^n x_j$

所以合并的时候（假设 ls 是左儿子，rs 是右儿子）：

$tree_p=tree_{ls}+tree_{rs}+num_{ls}\ast sum_{rs}+num_{rs}\ast sum_{ls}$

Code

具体写起来要考虑一(很)点(多)细节 :new_moon_with_face: 。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<vector>
#include<map>

using namespace std;
#define int long long

inline int read(){
	int ret=0,f=1;char ch=getchar();
	while (ch<'0'||ch>'9') {if (ch=='-') f=-1;ch=getchar();}
	while (ch>='0'&&ch<='9') ret=ret*10+ch-'0',ch=getchar();
	return ret*f;
}

const int maxn=2e5+5;
const pair<int,int> zero_pair=make_pair(0,0);
int n,m,N;
map<int,int> to;
 numto[maxn*],sx[maxn];





 {
	 x,y,z;
}zero_sts=(sts){,,};

 SegmentTree{
	 tree[maxn*],sum[maxn*],num[maxn*];

	{
		 (x==tl && tl==tr){
			sum[p]+=delta*numto[x];
			num[p]+=delta;
			;
		}
		 (x<=mid  ) (x,delta,tl,mid,ls);
		 (mid<=x) (x,delta,mid,tr,rs);
		num[p]=num[ls]+num[rs];
		tree[p]=tree[ls]+tree[rs]+num[ls]*sum[rs]-num[rs]*sum[ls];
		sum[p]=sum[rs]+sum[ls];
		
	}

	{
		 (sl<=tl && tr<=sr){
			
			 (sts){sum[p],tree[p],num[p]};
		}
		sts ret1=zero_sts,ret2=zero_sts;
		 (sl<=mid  ) ret1=(sl,sr,tl,mid,ls);
		 (mid<=sr) ret2=(sl,sr,mid,tr,rs);
		sts ret=(sts){retx+retx,  rety+rety+retz*retx-retz*retx,  retz+retz};
		
		 ret;
	}
}

 {
	 x,y,z;
}opts[maxn];

{
	n=();
	vector<> vec;vec.();
	 ( i=;i<=n;i++) numto[i]=sx[i]=(),vec.(numto[i]);
	m=();
	 ( i=;i<=m;i++){
		opts[i].x=(),opts[i].y=(),opts[i].z=();
		 (opts[i].x==) numto[opts[i].y]+=opts[i].z,vec.(numto[opts[i].y]);
		 vec.(opts[i].y),vec.(opts[i].z);
	}
	(vec.(),vec.());(vec.(),vec.()); N=vec.();
	 ( i=;i<=N;i++) numto[i]=vec[i],to[numto[i]]=i;
	
	 ( i=;i<=n;i++) SegmentTree::(to[sx[i]],,,N,);
	 ( i=;i<=m;i++){
		 (opts[i].x==){
			SegmentTree::(to[sx[opts[i].y]],,,N,);
			sx[opts[i].y]+=opts[i].z;
			SegmentTree::(to[sx[opts[i].y]],,,N,);
		}  {
			sts ans=SegmentTree::(to[opts[i].y],to[opts[i].z],,N,);
			(,ans.y);
			
		}
	}
	 ;
}

1/30/2019