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;
}
}
}

}