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

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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 $x_{p_j} + d_j$ . 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 $[l_j, r_j] (l_j ≤ r_j)$ . In other words, you should count the sum of: $\displaystyle \sum_{l_j\leq x_p\leq x_q\leq r_j} (x_q-x_p)$

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 ( $|x_i| ≤ 10^9$ ).

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 $t_j$ ( $1 ≤ t_j ≤ 2$ ) — the query type. If $t_j = 1$ , then it is followed by two integers $p_j$ and $d_j$ ( $1 ≤ p_j ≤ n, |d_j| ≤ 1000$ ). If $t_j = 2$ , then it is followed by two integers $l_j$ and $r_j$ ( $- 10^9 ≤ l_j ≤ r_j ≤ 10^9$ ).

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 $\sum_{x_i>x_j} x_i-x_j$ ，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 $\sum_{x_i>x_j} x_i-x_j$ ；
$num_p$ ：这个区间中点的数量。接下来需要考虑如何合并。对于 $sum_p$ 和 $num_p$ 都可以子树对应直接相加，只有 $tree_p$ 比较麻烦：

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

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;
int numto[maxn*2],sx[maxn];

#define ls ((p<<1))
#define rs ((p<<1)+1)
#define mid (((tr-tl)>>1)+tl)

struct sts{
	int x,y,z;
}zero_sts=(sts){0,0,0};

namespace SegmentTree{
	int tree[maxn*4],sum[maxn*4],num[maxn*4];

	inline void change(int x,int delta,int tl,int tr,int p){
		if (x==tl && tl==tr){
			sum[p]+=delta*numto[x];
			num[p]+=delta;
			return;
		}
		if (x<=mid  ) change(x,delta,tl,mid,ls);
		if (mid+1<=x) change(x,delta,mid+1,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];
		// printf("[%lld,%lld]: sum=%lld tree=%lld num=%lld\n",tl,tr,sum[p],tree[p],num[p]);
	}

	inline sts query(int sl,int sr,int tl,int tr,int p){
		if (sl<=tl && tr<=sr){
			// printf("tl=%lld tr=%lld sum=%lld tree=%lld num=%lld\n",tl,tr,sum[p],tree[p],num[p]);
			return (sts){sum[p],tree[p],num[p]};
		}
		sts ret1=zero_sts,ret2=zero_sts;
		if (sl<=mid  ) ret1=query(sl,sr,tl,mid,ls);
		if (mid+1<=sr) ret2=query(sl,sr,mid+1,tr,rs);
		sts ret=(sts){ret1.x+ret2.x,  ret1.y+ret2.y+ret1.z*ret2.x-ret2.z*ret1.x,  ret1.z+ret2.z};
		// printf("query (%lld %lld %lld %lld) : sum=%lld tree=%lld num=%lld\n",sl,sr,tl,tr,ret.x,ret.y,ret.z);;
		return ret;
	}
}

struct opt{
	int x,y,z;
}opts[maxn];

signed main(){
	n=read();
	vector<int> vec;vec.clear();
	for (int i=1;i<=n;i++) numto[i]=sx[i]=read(),vec.push_back(numto[i]);
	m=read();
	for (int i=1;i<=m;i++){
		opts[i].x=read(),opts[i].y=read(),opts[i].z=read();
		if (opts[i].x==1) numto[opts[i].y]+=opts[i].z,vec.push_back(numto[opts[i].y]);
		else vec.push_back(opts[i].y),vec.push_back(opts[i].z);
	}
	sort(vec.begin(),vec.end());unique(vec.begin(),vec.end()); N=vec.size();
	for (int i=1;i<=N;i++) numto[i]=vec[i-1],to[numto[i]]=i;//,cout<<i<<" --> "<<vec[i-1]<<endl;
	
	for (int i=1;i<=n;i++) SegmentTree::change(to[sx[i]],1,1,N,1);
	for (int i=1;i<=m;i++){
		if (opts[i].x==1){
			SegmentTree::change(to[sx[opts[i].y]],-1,1,N,1);
			sx[opts[i].y]+=opts[i].z;
			SegmentTree::change(to[sx[opts[i].y]],1,1,N,1);
		} else {
			sts ans=SegmentTree::query(to[opts[i].y],to[opts[i].z],1,N,1);
			printf("%lld\n",ans.y);
			// printf(" - %lld %lld %lld\n",ans.x,ans.y,ans.z);
		}
	}
	return 0;
}

1/30/2019