算法随笔——分块
介绍
分块的基本思想是通过适当的划分和预处理，用空间换时间，更加接近朴素算法，是一种暴力数据结构。
例题1
例如最经典的区间修改区间查询，若用树状数组来做就显得过于麻烦了。而用线段树做这道题，虽然通用，但马亮比较大，非常不友好。于是考虑分块。
思路
将整个序列分为 $\sqrt{n} $ 块，每个块长自然为 \(\sqrt{n}\)，于是采用大段维护，小段朴素的思路。在查询或修改的范围 \([l,r]\) 包含完整的块时，直接通过 add 数组标记该区间整体的变化。而左右两端零散的块直接朴素枚举，肥肠暴力。
代码
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define int ll
const int N = 2e5+5;
int n,m,a[N],L[N],R[N],pos[N],sum[N],add[N];
void modify(int l,int r,int d)
{
int p = pos[l],q = pos[r];
if (p == q)
{
sum[p] += (r-l+1) * d;
for (int i = l;i <= r;i++) a[i] += d;
return;
}
for (int i = p+1;i <= q-1;i++) add[i] += d;
for (int i = l ;i <= R[p];i++) a[i] += d;
for (int i = L[q];i <= r;i++) a[i] += d;
sum[p] += (R[p]-l+1) * d;
sum[q] += (r-L[q]+1) * d;
}
int query(int l,int r)
{
int p = pos[l],q = pos[r];
ll ans = 0;
if (p == q)
{
for (int i = l;i <= r;i++) ans += a[i] + add[p];
return ans;
}
for (int i = p + 1;i <= q-1;i++) ans += sum[i] + add[i] * (R[i]-L[i]+1);
for (int i = l;i <= R[p];i++) ans += a[i] + add[p];
for (int i = L[q];i <= r;i++) ans += a[i] + add[q];
return ans;
}
signed main()
{
cin >> n >> m;
for (int i = 1;i <= n;i++) cin >> a[i];
int t = sqrt(n);
for (int i = 1;i <= t;i++)
L[i] = (i-1) * t + 1,R[i] = i * t;
if (R[t] < n) t++,R[t] = n,L[t] = R[t-1] + 1;
for (int i = 1;i <= t;i++)
for (int j = L[i];j <= R[i];j++)
pos[j] = i,sum[i] += a[j];
for (int i = 1;i <= m;i++)
{
char op[2];
scanf("%s",op);
int l,r,d;
if (op[0] == 'Q')
{
cin >> l >> r;
cout << query(l,r) << endl;
}
else
{
cin >> l >> r >> d;
modify(l,r,d);
}
}
return 0;
}
例题2
一句话题意：给出 \(l,r\)，求出 \([l,r]\) 的区间众数。
本题在线求解区间众数，众数不满足区间可加性，因此若用线段树和树状数组做较难维护，所以考虑分块。
做法
同样采用分块思想，将原序列分成 \(T\) 块，则块长为 $L = \frac{N}{T} $。
-
预处理每个区间 \([L,R]\) 中每个数出现的个数。这里的 \(L,R\) 是块的边界，一共有 \(T\) 块，所以时间复杂度为 $O(T ^2) $。同时每个区间都需要空间为 \(O(N)\) 的数组存储每个数出现的个数，空间复杂度为 \(O(NT^2)\)，记为 \(cnt_{L,R}\)。
-
而对于不在块中的查询范围 \([l,L],[R,r]\) 可以用朴素扫描，在预处理好的 \([L,R]\) 基础上累加，时间复杂度为 \(O(Q \times \frac{N}{T})\)。
设 \(Q\) 与 \(N\) 数量级相等，最终的时间复杂度为 \(O(NT^2+\frac{N^2}{T})\)。
根据基本不等式：
\[\begin{aligned} NT^2+\frac{N^2}{T} &\ge 2 \times \sqrt{NT^2 \cdot \frac{N^2}{T}} \\ &= 2\sqrt{TN^3}
\end{aligned}\]
当且仅当 \(\large NT^2 = \frac{N^2}{T}\)，即 $\large T = \sqrt[3]{N} $ 时，原式取得最小值。
因此该算法时间复杂度为 \(\large O(N ^ {\frac{3}{5}})\)，可以通过本题。
完整代码
点击查看代码
// Problem: 钂插叕鑻?
// Contest: AcWing
// URL: https://www.acwing.com/problem/content/251/
// Memory Limit: 256 MB
// Time Limit: 3000 ms
// Author: Eason
// Date:2024-01-18 09:47:03
//
// Powered by CP Editor (https://cpeditor.org)
#include<bits/stdc++.h>
using namespace std;
#define ull unsigned long long
#define ll long long
#define INF 0x3f3f3f3f
#define re register
#define il inline
#define gc getchar
inline int read()
{
int f=1,k=0;
char c = getchar();
while (c <'0' || c > '9')
{
if (c=='-') f=-1;
c=getchar();
}
while(c >= '0' && c <= '9') k = (k << 1)+(k << 3)+(c^48),c=getchar();
return k*f;
}
const int N = 50005,M = 40;
int n,m,a[N];
int L[N],R[N],pos[N];
int block[M][M][N],mc[M][M],mck[M][M];
unordered_map<int,int> mp,timp;
int idx;
int main()
{
cin >> n >> m;
vector<int> v;
for (int i = 1;i <= n;i++)
{
cin >> a[i];
v.push_back(a[i]);
}
sort(v.begin(),v.end());
v.erase(unique(v.begin(),v.end()),v.end());
for (int i = 0;i < v.size();i++) mp[v[i]] = i;
for (int i = 1;i <= n;i++) a[i] = mp[a[i]];
int T = cbrt(n),len = n/T;
for (int i = 1;i <= T;i++)
L[i] = (i-1) * len+1,R[i] = i * len;
if (R[T] < n) T++,R[T] = n,L[T] = R[T-1] + 1;
for (int i = 1;i <= T;i++)
for (int j = L[i];j <= R[i];j ++)
pos[j] = i;
for (int i = 1;i <= T;i++)
for (int j = i;j <= T;j++)
{
int maxn = -1e9,maxk;
for (int k = L[i];k <= R[j];k++)
{
block[i][j][a[k]] ++;
if (block[i][j][a[k]] > maxn) maxn = block[i][j][a[k]],maxk = a[k];
else if (block[i][j][a[k]] == maxn)
{
if (a[k] <= maxk) maxn = block[i][j][a[k]],maxk = a[k];
}
}
mc[i][j] = maxn;
mck[i][j] = maxk;
}
int x = 0;
for (int i = 1;i <= m;i++)
{
int le,ri;
cin >> le >> ri;
le = (le +x-1) % n + 1,ri = (ri+x-1) % n + 1;
if (le > ri) swap(le,ri);
int pl = pos[le],pr = pos[ri],csmc = mc[pl+1][pr-1],csmck = mck[pl+1][pr-1];
if (pl == pr)
{
unordered_map<int,int> mp;
int maxn = -1e9,maxk = -1;
for (int j = le;j <= ri;j++) mp[a[j]]++;
for (int j = le;j <= ri;j++) if (mp[a[j]] > maxn || (mp[a[j]] == maxn && a[j] < maxk)) maxn = mp[a[j]],maxk = a[j];
cout << v[maxk] << endl;
x= v[maxk];
mp.clear();
continue;
}
for(int k = le;k <= R[pl];k++)
{
block[pl+1][pr-1][a[k]]++;
if (block[pl+1][pr-1][a[k]] > mc[pl+1][pr-1]) mc[pl+1][pr-1] = block[pl+1][pr-1][a[k]],mck[pl+1][pr-1] = a[k];
else if (block[pl+1][pr-1][a[k]] == mc[pl+1][pr-1])
{
if (a[k] <= mck[pl+1][pr-1]) mc[pl+1][pr-1] = block[pl+1][pr-1][a[k]],mck[pl+1][pr-1] = a[k];
}
}
for (int k = L[pr];k <= ri;k++)
{
block[pl+1][pr-1][a[k]]++;
if (block[pl+1][pr-1][a[k]] > mc[pl+1][pr-1]) mc[pl+1][pr-1] = block[pl+1][pr-1][a[k]],mck[pl+1][pr-1] = a[k];
else if (block[pl+1][pr-1][a[k]] == mc[pl+1][pr-1])
{
if (a[k] <= mck[pl+1][pr-1]) mc[pl+1][pr-1] = block[pl+1][pr-1][a[k]],mck[pl+1][pr-1] = a[k];
}
}
cout << (x = v[mck[pl+1][pr-1]]) << endl;
for(int k = le;k <= R[pl];k++)
block[pl+1][pr-1][a[k]]--;
for (int k = L[pr];k <= ri;k++)
block[pl+1][pr-1][a[k]]--;
mc[pl+1][pr-1] = csmc,mck[pl+1][pr-1] = csmck;
}
return 0;
}
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