线段树模板题
所以，我偏不用线段树
奇了怪了
主要思路：平衡树——Splay
Splay是可以很好的维护区间的。
我这里主要讲如何用Splay维护区间。
我们知道Splay是严格按照中序遍历的顺序的，用rotate操作并不会改变这种性质，所以我们我们可以考虑一下一棵二叉树的中序遍历的特点。
如果我们把左端点splay到树根，把右端点splay到树根的右儿子位置，我们再做下中序遍历，，，（可以自行脑补）
是不是根的右儿子的左子树的信息就是这段区间的信息？
所以我们用Splay维护区间时我们是提取区间。
完整代码：
（福利：其中还有插入节点，删除节点等操作哦QwQ，附带节点垃圾桶，可回收旧结点）
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
using namespace std;
#define go(i, j, n, k) for (int i = j; i <= n; i += k)
#define fo(i, j, n, k) for (int i = j; i >= n; i -= k)
#define rep(i, x) for (int i = h[x]; i; i = e[i].nxt)
#define mn 100010
#define ld long double
#define fi first
#define se second
#define inf 1 << 30
#define ll long long
#define root 1, n, 1
#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1
inline ll read()
{
ll x = 0, f = 1;
char ch = getchar();
while (ch > '9' || ch < '0')
{
if (ch == '-')
f = -f;
ch = getchar();
}
while (ch >= '0' && ch <= '9')
{
x = x * 10 + ch - '0';
ch = getchar();
}
return x * f;
}
vector<int> ljt;
struct tree
{
int ch[2], sze, fa;
ll w, sum, col;
tree(int _sze = 0, int _fa = 0, int _w = 0, int _sum = 0, int _col = 0)
: sze(_sze), fa(_fa), w(_w), sum(_sum), col(_col) { ch[1] = ch[0] = 0; }
} z[mn];
int n, m, tot, a[mn];
inline int newnode(int v = 0)
{
int rt;
if (ljt.empty())
rt = ++tot;
else
rt = ljt.back(), ljt.pop_back();
z[rt].fa = z[rt].ch[0] = z[rt].ch[1] = 0;
z[rt].sze = 1;
z[rt].w = z[rt].sum = v;
return rt;
}
inline void deletenode(int rt)
{
z[rt].fa = z[rt].ch[0] = z[rt].ch[1] = z[rt].w = z[rt].sum = 0;
z[rt].sze = 1;
ljt.push_back(rt);
}
inline void update(int rt)
{
z[rt].sum = z[rt].w, z[rt].sze = 1;
if(z[rt].ch[0])
z[rt].sum += z[z[rt].ch[0]].sum, z[rt].sze += z[z[rt].ch[0]].sze;
if(z[rt].ch[1])
z[rt].sum += z[z[rt].ch[1]].sum, z[rt].sze += z[z[rt].ch[1]].sze;
}
inline void push_col(int rt)
{
if(z[rt].col)
{
z[z[rt].ch[0]].col += z[rt].col;
z[z[rt].ch[1]].col += z[rt].col;
z[z[rt].ch[0]].sum += z[z[rt].ch[0]].sze * z[rt].col;
z[z[rt].ch[1]].sum += z[z[rt].ch[1]].sze * z[rt].col;
z[z[rt].ch[0]].w += z[rt].col;
z[z[rt].ch[1]].w += z[rt].col;
z[rt].col = 0;
}
}
inline int iden(int rt)
{
return z[z[rt].fa].ch[0] == rt ? 0 : 1;
}
inline void conn(int x, int y, int son)
{
z[x].fa = y;
z[y].ch[son] = x;
}
inline void rotate(int x)//敲好记的rotate函数！
{
int y = z[x].fa;
int moot = z[y].fa;
int yson = iden(x);
int mootson = iden(y);
int B = z[x].ch[yson ^ 1];
conn(B, y, yson), conn(y, x, yson ^ 1), conn(x, moot, mootson);
update(y), update(x);
}
inline void splay(int x, int &k)//传址要注意
{
if (x == k)
return;
int p = z[k].fa;
while (z[x].fa != p)
{
push_col(x);
int y = z[x].fa;
if (z[y].fa != p)
rotate(iden(y) ^ iden(x) ? x : y);
rotate(x);
}
k = x;
}
inline int findkth(int rt, int k)
{
while (233)
{
push_col(rt);
if (z[rt].ch[0] && k <= z[z[rt].ch[0]].sze)
rt = z[rt].ch[0];
else
{
if (z[rt].ch[0])
k -= z[z[rt].ch[0]].sze;
if (!--k)
return rt;
rt = z[rt].ch[1];
}
}
}
inline void insert(int &rt, int p, int v)//传址要注意
{
int x = findkth(rt, p);
splay(x, rt);
int y = findkth(rt, p + 1);
int ooo = z[rt].ch[1];
splay(y, ooo);
z[y].ch[0] = newnode(v);
z[z[y].ch[0]].fa = y;
update(z[y].ch[0]), update(y), update(x);
}
inline void erase(int &rt, int p)//传址要注意
{
int y = findkth(rt, p);
splay(y, rt);
int x = findkth(rt, p + 1);
int ooo = z[rt].ch[1];
splay(x, ooo);
int oo = z[x].ch[1];
z[oo].fa = y;
z[y].ch[1] = oo;
deletenode(x);
update(y);
}
inline int getRange(int &rt, int l, int r)//传址要注意
{
int x = findkth(rt, l);
splay(x, rt);
int y = findkth(rt, r + 2);
int ooo = z[rt].ch[1];
splay(y, ooo);
return z[y].ch[0];
}
inline void modify(int &rt, int l, int r, ll v)//传址要注意
{
int x = getRange(rt, l, r);
z[x].col += v;
z[x].w += v;
z[x].sum += z[x].sze * v;
update(z[rt].ch[1]);
update(rt);
}
inline ll query(int &rt, int l, int r)//传址要注意
{
int x = getRange(rt, l, r);
return z[x].sum;
}
inline void build(int rt, int l, int r)
{
int m = (l + r) >> 1;
z[rt].w = a[m];
if (l <= m - 1)
{
z[rt].ch[0] = newnode();
z[z[rt].ch[0]].fa = rt;
build(z[rt].ch[0], l, m - 1);
}
if (m + 1 <= r)
{
z[rt].ch[1] = newnode();
z[z[rt].ch[1]].fa = rt;
build(z[rt].ch[1], m + 1, r);
}
update(rt);
}
inline void debug(int rt)//debug专用，利用中序遍历
{
//if(!z[rt].ch[0] && !z[rt].ch[1])
// return;
if (rt == 0)
return;
debug(z[rt].ch[0]);
printf("%d %d %d
", z[rt].w, z[rt].sum, z[rt].sze);
debug(z[rt].ch[1]);
}
int main()
{
n = read();
m = read();
go(i, 1, n, 1) a[i] = read();
int rot = ++tot;
build(rot, 0, n + 1);
//debug(rot);
//cout << query(rot, 1, n) << "
";
go(i, 1, m, 1)
{
int s = read(), x = read(), y = read();
if (s == 1)
{
int v = read();
modify(rot, x, y, v);
}
else
{
cout << query(rot, x, y) << "
";
}
}
return 0;
}