支持访问任意历史版本以及在历史版本上修改的数据结构 每次修改不是在原有线段树上直接修改，而是在旁边新建点。
时空复杂度都是O(mlogn)
可持久化线段树

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const int N = 100000 + 5;//点数
long long tree[N * 25], add_num[N * 25];
int total, left_child[N * 25], right_child[N * 25], root[N * 25];
/****************************************************************************
逻辑上仍然是完全二叉树，动态加点，所以左右儿子不能再通过*2或*2+1得到
tree[]为存放树的(答案)，left_child[t]\right_child[t]存放t的左右儿子在tree中的下标
total用于加点时分配下标，root用于更新时创建新的根，不同的root为不同的时间
add_num[]存放每个节点区间更新的标记
*****************************************************************************/
void build(int t, int l, int r)//t为在tree中的下标，l为当前递归状态表示的l，r为当前r
{
	if (l == r)
	{
		//tree[t]=a[l];
		//cin >> tree[t];
		//向叶子节点添加值
		return;
	}
	int mid = l + r >> 1;
	build(left_child[t] = ++total, l, mid);//向左递归时加一个点
	build(right_child[t] = ++total, mid + 1, r);//向右递归同时加一个点为右儿子
	tree[t] = tree[left_child[t]] + tree[right_child[t]];
	//tree[t] = max(tree[left_child[t]] , tree[right_child[t]]);
	//回溯的时候更新父亲，以求区间和为例
}


/**************************参数****************************
l,r同上，因为是新建的一条链，所以last和cur与上面的t意义一样，只是
在这里分成两个，一个新的一个旧的。x,y为查询区间,k为区间要加的东西
***********************************************************/
void interval_update(int last, int cur, int l, int r, int x, int y, int k)
{
	tree[cur] = tree[last] + (y-x + 1)*k;//更新当前区间
	right_child[cur] = right_child[last];//初始化这个新建的点，新点指向原先的左右儿子
	left_child[cur] = left_child[last];
	add_num[cur] = add_num[last];//原先这个点的标记拿到新建的这个点
	if (x <= l&&r <= y)
	{
		add_num[cur] += k;//更新，当前区间打上标记
		return;
	}
	int mid = l + r >> 1;
	if (y <= mid)
	{
		interval_update(left_child[last], left_child[cur] = ++total, l, mid, x, y, k);
	}
	else if (x>mid)
	{
		interval_update(right_child[last], right_child[cur] = ++total, mid + 1, r, x, y, k);
	}
	else
	{
		//往两边递归的时候查询的区间也要跟着分开
		interval_update(left_child[last], left_child[cur] = ++total, l, mid, x, mid, k);
		interval_update(right_child[last], right_child[cur] = ++total, mid + 1, r, mid+1, y, k);
	}
}
long long  interval_query(int t, int l, int r, int x, int y, int sum)//t,l,r,x,y同上，sum为一路传下来的标记
{
	if (x <= l&&r <= y)
	{
		//返回当前点的值和所有传下来的标记
		return tree[t] + (y - x + 1)*sum;
	}
	int mid = l + r >> 1;
	if (y <= mid)//向左递归
	{
		return interval_query(left_child[t], l, mid, x, y, sum + add_num[t]);
	}
	else if (x>mid)//向右递归
	{
		return interval_query(right_child[t], mid + 1, r, x, y, sum + add_num[t]);
	}
	else//区间分两半
	{
		//往两边递归的时候查询的区间也要跟着分开
		return interval_query(left_child[t], l, mid, x, mid, sum + add_num[t]) + interval_query(right_child[t], mid + 1, r, mid+1, y, sum + add_num[t]);
	}
}

例题

HDU4348

Problem Description

Background
To The Moon is a independent game released in November 2011, it is a role-playing adventure game powered by RPG Maker.
The premise of To The Moon is based around a technology that allows us to permanently reconstruct the memory on dying man. In this problem, we’ll give you a chance, to implement the logic behind the scene.

You‘ve been given N integers A[1], A[2],…, A[N]. On these integers, you need to implement the following operations:

  1. C l r d: Adding a constant d for every {Ai | l <= i <= r}, and increase the time stamp by 1, this is the only operation that will cause the time stamp increase.
  2. Q l r: Querying the current sum of {Ai | l <= i <= r}.
  3. H l r t: Querying a history sum of {Ai | l <= i <= r} in time t.
  4. B t: Back to time t. And once you decide return to a past, you can never be access to a forward edition anymore.
    … N, M ≤ 105, |A[i]| ≤ 109, 1 ≤ l ≤ r ≤ N, |d| ≤ 104 … the system start from time 0, and the first modification is in time 1, t ≥ 0, and won’t introduce you to a future state.
Input

n m
A1 A2 … An
… (here following the m operations. )

Output

… (for each query, simply print the result. )

Sample Input

10 5
1 2 3 4 5 6 7 8 9 10
Q 4 4
Q 1 10
Q 2 4
C 3 6 3
Q 2 4

2 4
0 0
C 1 1 1
C 2 2 -1
Q 1 2
H 1 2 1

Sample Output

4
55
9
15

0
1

题意：四种操作：
C L R D: 对l-r都加d，时间加一
q l r：询问l-r的总和
h l r t：查询t时刻l-r的总和
b t： 回到t
时间从1开始

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#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
using namespace std;
const int N = 100000 + 5;
long long tre[N * 25], add[N * 25];
int tot, lc[N * 25], rc[N * 25], root[N * 25];
void build(int t, int l, int r)
{
	if (l == r)
	{
		//		tre[t]=a[l];
		cin >> tre[t];
		return;
	}
	int mid = l + r >> 1;
	build(lc[t] = ++tot, l, mid);
	build(rc[t] = ++tot, mid + 1, r);
	tre[t] = tre[lc[t]] + tre[rc[t]];
}

void update(int last, int cur, int l, int r, int x, int y, int k)
{
	tre[cur] = tre[last] + (y-x + 1)*k;
	rc[cur] = rc[last];
	lc[cur] = lc[last];
	add[cur] = add[last];
	if (x <= l&&r <= y)
	{
		add[cur] += k;
		return;
	}
	int mid = l + r >> 1;
	if (y <= mid)
	{
		update(lc[last], lc[cur] = ++tot, l, mid, x, y, k);
	}
	else if (x>mid)
	{
		update(rc[last], rc[cur] = ++tot, mid + 1, r, x, y, k);
	}
	else
	{
		update(lc[last], lc[cur] = ++tot, l, mid, x, mid, k);
		update(rc[last], rc[cur] = ++tot, mid + 1, r, mid+1, y, k);
	}
}
long long  query(int t, int l, int r, int x, int y, int sum)
{
	if (x <= l&&r <= y)
	{
		return tre[t] + (y - x + 1)*sum;
	}
	int mid = l + r >> 1;
	if (y <= mid)
	{
		return query(lc[t], l, mid, x, y, sum + add[t]);
	}
	else if (x>mid)
	{
		return query(rc[t], mid + 1, r, x, y, sum + add[t]);
	}
	else
	{
		return query(lc[t], l, mid, x, mid, sum + add[t]) + query(rc[t], mid + 1, r, mid+1, y, sum + add[t]);
	}
}

int main()
{
	cout << log2(100000);
	int n, m;
	while (~scanf("%d%d", &n, &m))
	{
		memset(add, 0, sizeof(add));
		memset(lc, 0, sizeof(lc));
		memchr(rc, 0, sizeof(rc));
		tot = 0;
		build(root[1] = ++tot, 1, n);
		int time = 1;
		while (m--)
		{
			char s[10];
			scanf("%s", s);
			if (s[0] == 'C')
			{
				int l, r, k;
				scanf("%d%d%d", &l, &r, &k);
				time++;
				update(root[time - 1], root[time] = ++tot, 1, n, l, r, k);
			}
			else if (s[0] == 'Q')
			{
				int l, r;
				scanf("%d%d", &l, &r);
				cout << query(root[time], 1, n, l, r, 0) << endl;
			}
			else if (s[0] == 'H')
			{
				int l, r, t;
				cin >> l >> r >> t;
				cout << query(root[t + 1], 1, n, l, r, 0) << endl;
			}
			else if (s[0] == 'B')
			{
				int t;
				cin >> t;
				time = t + 1;
			}
		}
	}

}