注意事项:

  1. 分块可以看作一个两层的线段树。

代码:

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
void build(int n, int *a) {
sizb = sqrt(n);
int l = 1, r = sizb;
while (l <= n) {
le[++cntb] = l;
ri[cntb] = r;
len[cntb] = r - l + 1;
for (int i = l; i <= r; ++i) {
inb[i] = cntb;
// ...
}
l = r + 1, r = min(l + sizb - 1, n);
}
}
void modify(int l, int r, int v) {
int inbl = inb[l], inbr = inb[r];
if (inbl == inbr) {
// ...
} else {
// ...
}
}
int query(int l, int r) {
int inbl = inb[l], inbr = inb[r];
int ans = 0;
if (inbl == inbr) {
// ...
} else {
// ...
}
return ans;
}

Luogu P2357 守墓人

不建议使用线段树的模板题测试分块代码,数据过水(不加「query -> if (inbl == inbr) -> ans += (r - l + 1) * lzaddb[inbl];」也能过)。

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
#include<bits/stdc++.h>
using namespace std;
#define int long long
const int MAXn = 2e5;
const int MAXsqrtn = 448;

template <typename T>
inline void read(T &a) {
char c;for (c = getchar(); (c < '0' || c > '9') && c != '-'; c = getchar());bool f = c == '-';T x = f ? 0 : (c ^ '0');for (c = getchar(); c >= '0' && c <= '9'; c = getchar()) {x = x * 10 + (c ^ '0');}a = f ? -x : x;
}
template <typename T, typename ...Argv>
inline void read(T &a, Argv &...argv) {
read(a), read(argv...);
}

int sizb, cntb, le[MAXsqrtn + 10], ri[MAXsqrtn + 10], len[MAXsqrtn + 10], inb[MAXn + 10];
int sumb[MAXsqrtn + 10], lzaddb[MAXsqrtn + 10];
int sums[MAXn + 10];
void build(int n, int *a) {
sizb = sqrt(n);
int l = 1, r = sizb;
while (l <= n) {
le[++cntb] = l;
ri[cntb] = r;
len[cntb] = r - l + 1;
for (int i = l; i <= r; ++i) {
inb[i] = cntb;
sumb[cntb] += a[i];
sums[i] = a[i];
}
l = r + 1, r = min(l + sizb - 1, n);
}
}
void modify(int l, int r, int v) {
int inbl = inb[l], inbr = inb[r];
if (inbl == inbr) {
sumb[inbl] += (r - l + 1) * v;
for (int i = l; i <= r; ++i) sums[i] += v;
} else {
sumb[inbl] += (ri[inbl] - l + 1) * v;
for (int i = l; i <= ri[inbl]; ++i) sums[i] += v;
sumb[inbr] += (r - le[inbr] + 1) * v;
for (int i = le[inbr]; i <= r; ++i) sums[i] += v;
for (int i = inbl + 1; i < inbr; ++i) {
sumb[i] += len[i] * v;
lzaddb[i] += v;
}
}
}
int query(int l, int r) {
int inbl = inb[l], inbr = inb[r];
int ans = 0;
if (inbl == inbr) {
for (int i = l; i <= r; ++i) ans += sums[i];
ans += (r - l + 1) * lzaddb[inbl];
} else {
for (int i = l; i <= ri[inbl]; ++i) ans += sums[i];
ans += (ri[inbl] - l + 1) * lzaddb[inbl];
for (int i = le[inbr]; i <= r; ++i) ans += sums[i];
ans += (r - le[inbr] + 1) * lzaddb[inbr];
for (int i = inbl + 1; i < inbr; ++i) ans += sumb[i];
}
return ans;
}

int n, m;
int a[MAXn + 10];
signed main() {
read(n, m);
for (int i = 1; i <= n; ++i) {
read(a[i]);
}
build(n, a);
for (int i = 1, opt, l, r, v; i <= m; ++i) {
read(opt);
if (opt == 1) {
read(l, r, v);
modify(l, r, v);
} else if (opt == 2) {
read(v);
sumb[1] += v;
sums[1] += v;
} else if (opt == 3) {
read(v);
sumb[1] -= v;
sums[1] -= v;
} else if (opt == 4) {
read(l, r);
printf("%lld\n", query(l, r));
} else {
printf("%lld\n", lzaddb[1] + sums[1]);
}
}
return 0;
}