数列分块入门1-9 LibreOJ
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数列分块入门1-9 LibreOJ
我也不知道为什么一个大二的ACM选手没学分块。
我怎么记得大一的时候,学长教给我的分块就只有 block 和 num 两个变量来着...好吧,应该是我没认真学。正好前两天朋友给学弟开课,乘机去蹭了一节课。然后...我还是不会哇,菜的一逼塌糊涂。
还是卿学姐好哇,多听几遍,睡得贼香。
分块原理
分块嘛,其实就是优雅的暴力,和莫队(不会)有点异曲同工的赶脚。通过将数组分成小块以降低复杂度。
通常情况下:
- 每个块的大小(block)为 \(\sqrt{n}\)
- 块数(num)为 \(\sqrt{n}\)或\(\sqrt{n}+1\)
- 每个块的左端点(L[x])为 \((i-1)*block+1\)
- 每个块的右端点(R[x])为 \(i*block\)(最后一块右端点为\(n\))
- 位置 x 上的数属于(belong[x])第 \((i-1)/block+1\) 块
每次修改或查询的时候如果需要维护 [L, R] 的数,如果L、R被分在了同一块,那么就直接暴力跑就行了,反正复杂度不会超过\(\sqrt{n}\)。
如果不在一块,那就先暴力更新所有和 L 同一块且位置在 L 之后的数,再更新所有和 R 在同一块且位置在 R 之前的数,最后,再去把这中间的若干块直接按“块”更新(利用新数组之类的)。这样的复杂度也只有 \(\sqrt{n}\)量级。从而完成多次优雅的在线更新和查询。
LOJ 6277 区间加法更新 单点查询
没啥好些的,就按照上边写的来就行。因为是纯板子题,所以还是放个代码好了。
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 5e4 + 5;
int belong[maxn];
int block, num;
int l[maxn], r[maxn];
int n;
int a[maxn];
int more[maxn];
void build() {
block = sqrt(n);
num = n / block;
if (n % block != 0) {
num++;
}
for (int i = 1; i <= num; i++) {
l[i] = (i - 1) * block + 1;
r[i] = i * block;
}
r[num] = n;
for (int i = 1; i <= n; i++) {
belong[i] = (i - 1) / block + 1;
}
}
void update(int l, int r, int c) {
if (belong[l] == belong[r]) {
for (int i = l; i <= r; i++) {
a[i] += c;
}
return;
}
for (int i = l; belong[i] == belong[l]; i++) {
a[i] += c;
}
for (int i = r; belong[i] == belong[r]; i--) {
a[i] += c;
}
for (int i = belong[l] + 1; i <= belong[r] - 1; i++) {
more[i] += c;
}
}
int query(int x) {
return a[x] + more[belong[x]];
}
int main() {
while (~scanf("%d", &n)) {
memset(more, 0, sizeof(more));
memset(belong, 0, sizeof(belong));
memset(a, 0, sizeof(a));
for (int i = 1; i <= n; i++) {
scanf("%d", &a[i]);
}
build();
for (int i = 1; i <= n; i++) {
int f, l, r, c;
scanf("%d%d%d%d", &f, &l, &r, &c);
if (f == 0) {
update(l, r, c);
} else {
printf("%d\n", query(r));
}
}
}
return 0;
}
LOJ 6278 区间加法更新 区间查询
需要 vector 数组进行区间排序,然后 lower_bound 二分找答案就行了。
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 5e4 + 5;
int belong[maxn];
int block, num;
int l[maxn], r[maxn];
int a[maxn];
int more[maxn];
vector<int> v[maxn];
int n;
void build() {
block = sqrt(n);
num = n / block;
if (n % block != 0) {
num++;
}
for (int i = 1; i <= num; i++) {
l[i] = (i - 1) * block + 1;
r[i] = i * block;
}
r[num] = n;
for (int i = 1; i <= n; i++) {
belong[i] = (i - 1) / block + 1;
v[belong[i]].push_back(a[i]);
}
for (int i = 1; i <= num; i++) {
sort(v[i].begin(), v[i].end());
}
}
void pushup(int x) {
v[x].clear();
for (int i = l[x]; i <= r[x]; i++) {
v[x].push_back(a[i]);
}
sort(v[x].begin(), v[x].end());
}
void update(int l, int r, int c) {
if (belong[l] == belong[r]) {
for (int i = l; i <= r; i++) {
a[i] += c;
}
pushup(belong[l]);
return;
}
for (int i = l; belong[i] == belong[l]; i++) {
a[i] += c;
}
pushup(belong[l]);
for (int i = r; belong[i] == belong[r]; i--) {
a[i] += c;
}
pushup(belong[r]);
for (int i = belong[l] + 1; i <= belong[r] - 1; i++) {
more[i] += c;
}
}
int query(int l, int r, int c) {
int ans = 0;
if (belong[l] == belong[r]) {
for (int i = l; i <= r; i++) {
if (a[i] + more[belong[i]] < c) {
ans++;
}
}
return ans;
}
for (int i = l; belong[i] == belong[l]; i++) {
if (a[i] + more[belong[i]] < c) {
ans++;
}
}
for (int i = r; belong[i] == belong[r]; i--) {
if (a[i] + more[belong[i]] < c) {
ans++;
}
}
for (int i = belong[l] + 1; i <= belong[r] - 1; i++) {
int temp = c - more[i];
ans += lower_bound(v[i].begin(), v[i].end(), temp) - v[i].begin();
}
return ans;
}
int main() {
while (~scanf("%d", &n)) {
memset(more, 0, sizeof(more));
memset(belong, 0, sizeof(belong));
for (int i = 1; i <= n; i++) {
scanf("%d", &a[i]);
}
build();
for (int i = 1; i <= n; i++) {
int f, l, r;
int c;
scanf("%d%d%d%d", &f, &l, &r, &c);
if (f == 0) {
update(l, r, c);
} else {
printf("%d\n", query(l, r, c * c));
}
}
}
return 0;
}
LOJ 6279 区间加法更新 区间查询
几乎和第二题没差,就不上代码了。
LOJ 6280 区间加法更新 区间查询
几乎和第一题没差,需要维护每个块的和 s 数组,查询区间和。
LOJ 6281 区间开方更新 区间查询
这题相对前几题来说要难上不少,刚开始做的时候没反应过来,后来开方操作最多只会做64次,那么就肯定可以暴力了,不过需要标记整个块里是不是只有 0或1 (不需要开方操作了)。然后就是正常的区间加法,毕竟分块就是暴力哇。
#include <bits/stdc++.h>
using namespace std;
const int maxn = 5e4 + 5;
int belong[maxn];
int L[maxn], R[maxn];
int a[maxn];
int n;
int block, num;
int s[maxn];
int mark[maxn]; // 标记 1 为区间中只有 0 或 1
void build() {
block = sqrt(n);
num = n / block;
if (n % block) {
num++;
}
for (int i = 1; i <= num; i++) {
L[i] = (i - 1) * block + 1;
R[i] = i * block;
}
R[num] = n;
for (int i = 1; i <= n; i++) {
belong[i] = (i - 1) / block + 1;
s[belong[i]] += a[i];
}
}
void update(int l, int r) {
if (belong[l] == belong[r]) {
if (mark[belong[l]] == 0) {
mark[belong[l]] = 1;
for (int i = l; i <= r; i++) {
s[belong[i]] -= a[i];
a[i] = sqrt(a[i]);
s[belong[i]] += a[i];
}
for (int i = L[belong[l]]; i <= R[belong[l]]; i++) {
if (a[i] > 1)
mark[belong[l]] = 0;
}
}
return;
}
if (mark[belong[l]] == 0) {
mark[belong[l]] = 1;
for (int i = l; belong[i] == belong[l]; i++) {
s[belong[i]] -= a[i];
a[i] = sqrt(a[i]);
s[belong[i]] += a[i];
}
for (int i = L[belong[l]]; i <= R[belong[l]]; i++) {
if (a[i] > 1)
mark[belong[l]] = 0;
}
}
if (mark[belong[r]] == 0) {
mark[belong[r]] = 1;
for (int i = r; belong[i] == belong[r]; i--) {
s[belong[i]] -= a[i];
a[i] = sqrt(a[i]);
s[belong[i]] += a[i];
}
for (int i = L[belong[r]]; i <= R[belong[r]]; i++) {
if (a[i] > 1)
mark[belong[r]] = 0;
}
}
for (int i = belong[l] + 1; i <= belong[r] - 1; i++) {
if (mark[i] == 1)
continue;
s[i] = 0;
mark[i] = 1;
for (int j = L[i]; j <= R[i]; j++) {
a[j] = sqrt(a[j]);
s[i] += a[j];
if (a[j] > 1)
mark[i] = 0;
}
}
}
int query(int l, int r) {
int ans = 0;
if (belong[l] == belong[r]) {
for (int i = l; i <= r; i++) {
ans += a[i];
}
return ans;
}
for (int i = l; belong[i] == belong[l]; i++) {
ans += a[i];
}
for (int i = r; belong[i] == belong[r]; i--) {
ans += a[i];
}
for (int i = belong[l] + 1; i <= belong[r] - 1; i++) {
ans += s[i];
}
return ans;
}
int main() {
while (~scanf("%d", &n)) {
memset(mark, 0, sizeof(mark));
memset(belong, 0, sizeof(belong));
memset(s, 0, sizeof(s));
memset(a, 0, sizeof(a));
for (int i = 1; i <= n; i++) {
scanf("%d", &a[i]);
}
build();
for (int i = 1; i <= n; i++) {
int f, l, r, c;
scanf("%d%d%d%d", &f, &l, &r, &c);
if (f == 0) {
update(l, r);
} else {
printf("%d\n", query(l, r));
}
}
}
return 0;
}
LOJ 6282 单点插入更新 单点查询
每次在 l 位置前插入 r 数值,其实这是个 rope 裸题,stl 天下第一。那既然这是个分块专题,就得用分块做。
这题的难点就是插入达到\(\sqrt{n}\)后进行重新分块,不然会超时(比如一直往这块里添加)。
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 5;
int n;
int belong[maxn];
int a[maxn];
int block, num;
vector<int> v[maxn];
int temp[maxn << 2];
int cnt;
void build() {
block = sqrt(n);
num = n / block;
if (n % block) {
num++;
}
for (int i = 1; i <= n; i++) {
belong[i] = (i - 1) / block + 1;
v[belong[i]].push_back(a[i]);
}
}
void re_build() {
cnt = 0;
int tot = 0;
for (int i = 1; i <= num; i++) {
for (auto it = v[i].begin(); it != v[i].end(); it++) {
temp[tot++] = *it;
}
v[i].clear();
}
block = sqrt(tot);
for (int i = 0; i < tot; i++) {
v[(i - 1) / block + 1].push_back(temp[i]);
}
num = tot / block;
if (tot % block)
num++;
}
void query(int x, int &q, int &t) {
for (int i = 1; i <= num; i++) {
if (x > (int)v[i].size()) {
x -= (int)v[i].size();
} else {
q = i;
t = x - 1;
return;
}
}
}
void update(int l, int r) {
cnt++;
int q = -1, t = -1;
query(l, q, t);
v[q].insert(v[q].begin() + t, r);
if (cnt == block) {
re_build();
}
return;
}
int main() {
while (~scanf("%d", &n)) {
cnt = 0;
for (int i = 0; i <= n; i++) {
v[i].clear();
}
for (int i = 1; i <= n; i++) {
scanf("%d", &a[i]);
}
build();
for (int i = 1; i <= n; i++) {
int f, l, r, c;
scanf("%d%d%d%d", &f, &l, &r, &c);
if (f == 0) {
update(l, r);
} else {
int q = -1, t = -1;
query(r, q, t);
printf("%d\n", v[q][t]);
}
}
}
return 0;
}
LOJ 6283 区间加法+乘法更新 单点查询
这题比较麻烦,但是不算难,就是写起来烦,需要多维护一个数组,乘法分配律就完事了。然后我 memset 赋 1 debug 了半小时...
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int mod = 10007;
const int maxn = 1e5 + 5;
int n;
int block, num;
ll a[maxn];
ll belong[maxn];
ll more[maxn];
ll s[maxn];
int L[maxn], R[maxn];
void build() {
block = sqrt(n);
num = n / block;
if (n % block) {
num++;
}
for (int i = 1; i <= num; i++) {
L[i] = (i - 1) * block + 1;
R[i] = i * block;
}
R[num] = n;
for (int i = 1; i <= n; i++) {
belong[i] = (i - 1) / block + 1;
}
}
void pushup(int x) {
for (int i = L[x]; i <= R[x]; i++) {
a[i] = (a[i] * s[x] + more[x]) % mod;
}
s[x] = 1;
more[x] = 0;
}
void update_add(int l, int r, ll c) {
if (belong[l] == belong[r]) {
pushup(belong[l]);
for (int i = l; i <= r; i++) {
a[i] = (a[i] + c) % mod;
}
return;
}
pushup(belong[l]);
for (int i = l; belong[i] == belong[l]; i++) {
a[i] = (a[i] + c) % mod;
}
pushup(belong[r]);
for (int i = r; belong[i] == belong[r]; i--) {
a[i] = (a[i] + c) % mod;
}
for (int i = belong[l] + 1; i <= belong[r] - 1; i++) {
more[i] = (more[i] + c) % mod;
}
}
void update_mul(int l, int r, ll c) {
if (belong[l] == belong[r]) {
pushup(belong[l]);
for (int i = l; i <= r; i++) {
a[i] = (a[i] * c) % mod;
}
return;
}
pushup(belong[l]);
for (int i = l; belong[i] == belong[l]; i++) {
a[i] = (a[i] * c) % mod;
}
pushup(belong[r]);
for (int i = r; belong[i] == belong[r]; i--) {
a[i] = (a[i] * c) % mod;
}
for (int i = belong[l] + 1; i <= belong[r] - 1; i++) {
more[i] = (more[i] * c) % mod;
s[i] = (s[i] * c) % mod;
}
}
ll query(int x) { return a[x] * s[belong[x]] + more[belong[x]]; }
int main() {
while (~scanf("%d", &n)) {
for (int i = 0; i <= n; i++) {
s[i] = 1;
}
memset(more, 0, sizeof(more));
for (int i = 1; i <= n; i++) {
scanf("%lld", &a[i]);
}
build();
for (int i = 1; i <= n; i++) {
int f, l, r;
ll c;
scanf("%d%d%d%lld", &f, &l, &r, &c);
if (f == 0) {
update_add(l, r, c);
} else if (f == 1) {
update_mul(l, r, c);
} else {
printf("%lld\n", query(r) % mod);
}
}
}
return 0;
}
LOJ 6284 区间查询后修改
先查询 [l, r] 中有多少 c ,在把中间的数都改成 c 。多开一个 vis 数组,如果块中的所有数都并非都一样,则置 vis[x] = -1,此时需要强制暴力找;否则 vis 数组记录的就是整个块中的数是多少,满足条件就直接 ans += block 。
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 5;
int belong[maxn];
int n, block, num;
int a[maxn];
int L[maxn], R[maxn];
int vis[maxn];
void build() {
block = sqrt(n);
num = n / block;
if (n % block) {
num++;
}
for (int i = 1; i <= num; i++) {
L[i] = (i - 1) * block + 1;
R[i] = i * block;
}
R[num] = n;
for (int i = 1; i <= n; i++) {
belong[i] = (i - 1) / block + 1;
}
}
void pushup(int x) {
if (vis[x] == -1)
return;
for (int i = L[x]; i <= R[x]; i++) {
a[i] = vis[x];
}
vis[x] = -1;
}
int query(int l, int r, int c) {
int ans = 0;
if (belong[l] == belong[r]) {
pushup(belong[l]);
for (int i = l; i <= r; i++) {
if (a[i] == c) {
ans++;
}
a[i] = c;
}
return ans;
}
pushup(belong[l]);
for (int i = l; belong[i] == belong[l]; i++) {
if (a[i] == c) {
ans++;
}
a[i] = c;
}
pushup(belong[r]);
for (int i = r; belong[i] == belong[r]; i--) {
if (a[i] == c) {
ans++;
}
a[i] = c;
}
for (int i = belong[l] + 1; i <= belong[r] - 1; i++) {
if (vis[i] == -1) {
for (int j = L[i]; j <= R[i]; j++) {
if (a[j] == c) {
ans++;
}
}
} else {
if (vis[i] == c) {
ans += block;
}
}
vis[i] = c;
}
return ans;
}
int main() {
while (~scanf("%d", &n)) {
memset(belong, 0, sizeof(belong));
memset(vis, -1, sizeof(vis));
for (int i = 1; i <= n; i++) {
scanf("%d", &a[i]);
}
build();
for (int i = 1; i <= n; i++) {
int l, r, c;
scanf("%d%d%d", &l, &r, &c);
printf("%d\n", query(l, r, c));
}
}
return 0;
}
LOJ 6285 区间查询众数
这题...就很麻烦了,需要先预处理出每个块内的众数,然后再用 upper_bound - lower_bound 去得到某个数的个数。总之说起来挺复杂的,看着代码理解一下还是可以的。
吐槽一下讨论版玄学,这啥数据啊,块的大小还得是 30 ,我代码都重构的不成样子了...
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 5;
int n;
int block, num;
int belong[maxn];
int a[maxn];
map<int, int> mp; // a[i] 是第几个出现的
int val[maxn]; // val[id] = a[i]
int mark[maxn / 30][maxn / 30]; // 预处理每个整块的众数
int cnt[maxn];
vector<int> v[maxn]; // 记录每个数出现的每一个位置
/*void build() {
}*/
inline int get_id(int l, int r, int x) {
return upper_bound(v[x].begin(), v[x].end(), r) - lower_bound(v[x].begin(), v[x].end(), l);
}
inline int query(int l, int r) {
int MAX = 0, tot = 0;
if (belong[l] == belong[r]) {
for (int i = l; i <= r; i++) {
int temp = get_id(l, r, a[i]);
if (temp > tot || (temp == tot && val[a[i]] < val[MAX])) {
tot = temp;
MAX = a[i];
}
}
return MAX;
}
for (int i = l; belong[i] == belong[l]; i++) {
int temp = get_id(l, r, a[i]);
if (temp > tot || (temp == tot && val[a[i]] < val[MAX])) {
tot = temp;
MAX = a[i];
}
}
for (int i = r; belong[i] == belong[r]; i--) {
int temp = get_id(l, r, a[i]);
if (temp > tot || (temp == tot && val[a[i]] < val[MAX])) {
tot = temp;
MAX = a[i];
}
}
int res = mark[belong[l] + 1][belong[r] - 1];
int temp = get_id(l, r, res);
if (temp > tot || (temp == tot && val[res] < val[MAX])) {
tot = temp;
MAX = res;
}
return MAX;
}
int main() {
scanf("%d", &n);
int id = 0;
block = 30;
num = n / block;
if (n % block) {
num++;
}
for (int i = 1; i <= n; i++) {
scanf("%d", &a[i]);
if (mp[a[i]] == 0) {
mp[a[i]] = ++id;
val[id] = a[i];
}
belong[i] = (i - 1) / block + 1;
a[i] = mp[a[i]];
v[a[i]].push_back(i);
}
for (int i = 1; i <= num; i++) {
memset(cnt, 0, sizeof(cnt));
int MAX = 0, tot = 0;
for (int j = (i - 1) * block + 1; j <= n; j++) {
int temp = ++cnt[a[j]];
if (temp > tot || (temp == tot && val[a[j]] < val[MAX])) {
tot = temp;
MAX = a[j];
}
mark[i][belong[j]] = MAX;
}
}
for (int i = 1; i <= n; i++) {
int l, r;
scanf("%d%d", &l, &r);
printf("%d\n", val[query(l, r)]);
}
return 0;
}
总结一下分块,那就是暴力哇,惹不起惹不起.jpg,溜了溜了。
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