模板可持久化线段树

Posted 2021-01-24 yanyiming10243247

如题，这是一个模板。。。

 1 #include <algorithm>
 2 #include <iostream>
 3 #include <cstring>
 4 #include <cstdio>
 5 #include <cctype>
 6 
 7 inline void read(int & x)
 8 {
 9     x = 0;
10     int k = 1;
11     char c = getchar();
12     while (!isdigit(c))
13         if (c == ‘-‘) c = getchar(), k = -1;
14         else c = getchar();
15     while (isdigit(c))
16         x = (x << 1) + (x << 3) + (c ^ 48),
17         c = getchar();
18     x *= k;
19 }
20 
21 #define ls t[u].l
22 #define rs t[u].r
23 
24 const int MAXN = 1010100;
25 
26 struct SGT
27 {
28     int l, r, w;
29 }t[MAXN << 4];
30 
31 int n, m, opt, old, loc, x, id = 0;
32 int val[MAXN], root[MAXN];
33 
34 inline void Pushup(int u) { t[u].w = t[ls].w + t[rs].w; }
35 
36 void Build(int u, int l, int r)
37 {
38     if (l == r) 
39     {
40         t[u].w = val[l];
41         ls = rs = -1;
42         return;
43     }
44     ls = ++id, rs = ++id;
45     int mid = (l + r) >> 1;
46     Build(ls, l, mid);
47     Build(rs, mid + 1, r);
48     Pushup(u);
49 }
50 
51 int Add(int pre, int l, int r, int p, int c, int befor)
52 {
53     int cur = ++id;
54     if (l == r)
55     {
56         t[cur].w = c;
57         return cur;
58     }
59     int mid = (l + r) >> 1;
60     t[cur].l = t[pre].l;
61     t[cur].r = t[pre].r;
62     //t[cur].w = t[pre].w - befor + c;
63     if (l < r)
64         if (p <= mid) t[cur].l = Add(t[pre].l, l, mid, p, c, befor);
65         else t[cur].r = Add(t[pre].r, mid + 1, r, p, c, befor);
66     return cur;
67 }
68 
69 int Query(int u, int l, int r, int q)
70 {
71     if (l == r) return t[u].w;
72     int mid = (l + r) >> 1;
73     if (q <= mid) return Query(t[u].l, l, mid, q);
74     if (q > mid) return Query(t[u].r, mid + 1, r, q);
75 }
76 
77 signed main()
78 {
79     read(n), read(m);
80     for (int i = 1; i <= n; ++i) read(val[i]);
81     Build(root[0] = ++id, 1, n);
82     for (int i = 1; i <= m; ++i)
83     {
84         read(old), read(opt), read(loc);
85         if (opt == 1) 
86             read(x), root[i] = Add(root[old], 1, n, loc, x, Query(root[old], 1, n, loc));
87         if (opt == 2)
88             root[i] = root[i - 1], printf("%d
", Query(root[old], 1, n, loc));
89     }
90     return 0;
91 }