数列分块入门 1
区间加 + 单点查询
#include <iostream> #include <cstdio> #include <cmath> using namespace std; const int N = 5e4 + 10; #define gc getchar() inline int read() { int x = 0; char c = gc; while(c < '0' || c > '9') c = gc; while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = gc; return x; } int A[N], Add[N], bel[N]; int n, block, cnt; void Sec_G(int x, int y, int w) { if(bel[x] == bel[y]) for(int i = x; i <= y; i ++) A[i] += w; else { for(int i = x; i <= bel[x] * block; i ++) A[i] += w; for(int i = (bel[y] - 1) * block + 1; i <= y; i ++) A[i] += w; } for(int i = bel[x] + 1; i < bel[y]; i ++) Add[i] += w; } int main() { n = read(); block = sqrt(n); for(int i = 1; i <= n; i ++) A[i] = read(); for(int i = 1; i <= n; i ++) bel[i] = (i - 1) / block + 1; if(n % block) cnt = n / block + 1; else cnt = n / block; int T = n; while(T --) { int opt = read(), l = read(), r = read(), c = read(); if(!opt) Sec_G(l, r, c); else cout << A[r] + Add[bel[r]] << endl; } return 0; }
数列分块入门 2
区间加法，询问区间内小于某个值 x 的元素个数
用B[]记录A[], B[]数组中为排好序的A[]的映射
那么每次可以对每一块进行二分查找
#include <iostream> #include <cstdio> #include <cmath> #include <algorithm> using namespace std; const int N = 5e4 + 10; #define gc getchar() inline int read() { int x = 0; char c = gc; while(c < '0' || c > '9') c = gc; while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = gc; return x; } int A[N], B[N], Add[N], bel[N]; int n, block, cnt; void Work_sort(int x) { int l = (x - 1) * block + 1, r = min(l + block - 1, n); for(int i = l; i <= r; i ++) B[i] = A[i]; sort(B + l, B + r + 1); } void Sec_G(int x, int y, int w) { if(bel[x] == bel[y]) { for(int i = x; i <= y; i ++) A[i] += w; Work_sort(bel[x]); } else { for(int i = x; i <= bel[x] * block; i ++) A[i] += w; Work_sort(bel[x]); for(int i = (bel[y] - 1) * block + 1; i <= y; i ++) A[i] += w; Work_sort(bel[y]); } for(int i = bel[x] + 1; i < bel[y]; i ++) Add[i] += w; } inline int Calc(int x, int w) { int l = (x - 1) * block + 1, r = min(l + block - 1, n), ret = 0; while(l <= r) { int mid = (l + r) >> 1; if(B[mid] + Add[x] < w) ret = mid, l = mid + 1; else r = mid - 1; } return ret ? (ret - (x - 1) * block) : 0; } inline int Sec_A(int x, int y, int w) { int ret(0); if(bel[x] == bel[y]) { for(int i = x; i <= y; i ++) if(A[i] + Add[bel[x]] < w) ret ++; return ret; } else { for(int i = x; i <= bel[x] * block; i ++) if(A[i] + Add[bel[x]] < w) ret ++; for(int i = (bel[y] - 1) * block + 1; i <= y; i ++) if(A[i] + Add[bel[y]] < w) ret ++; } for(int i = bel[x] + 1; i < bel[y]; i ++) ret += Calc(i, w); return ret; } int main() { n = read(); block = sqrt(n); for(int i = 1; i <= n; i ++) A[i] = read(); for(int i = 1; i <= n; i ++) bel[i] = (i - 1) / block + 1; if(n % block) cnt = n / block + 1; else cnt = n / block; for(int i = 1; i <= cnt; i ++) Work_sort(i); int T = n; while(T --) { int opt = read(), l = read(), r = read(), c = read(); if(!opt) Sec_G(l, r, c); else cout << Sec_A(l, r, c * c) << " "; } return 0; }
数列分块入门 3
区间加法，询问区间内小于某个值 x 的前驱
与2类似，二分查找
#include <iostream> #include <cstdio> #include <cmath> #include <algorithm> using namespace std; const int N = 1e5 + 10; const int oo = 999999999; #define gc getchar() inline int read() { int x = 0; char c = gc; while(c < '0' || c > '9') c = gc; while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = gc; return x; } int A[N], B[N] = {-1}, Add[N], bel[N]; int n, block, cnt; void Work_sort(int x) { int l = (x - 1) * block + 1, r = min(l + block - 1, n); for(int i = l; i <= r; i ++) B[i] = A[i]; sort(B + l, B + r + 1); } void Sec_G(int x, int y, int w) { if(bel[x] == bel[y]) { for(int i = x; i <= y; i ++) A[i] += w; Work_sort(bel[x]); } else { for(int i = x; i <= bel[x] * block; i ++) A[i] += w; Work_sort(bel[x]); for(int i = (bel[y] - 1) * block + 1; i <= y; i ++) A[i] += w; Work_sort(bel[y]); } for(int i = bel[x] + 1; i < bel[y]; i ++) Add[i] += w; } inline int Calc(int x, int w) { int l = (x - 1) * block + 1, r = min(l + block - 1, n), ret = 0; while(l <= r) { int mid = (l + r) >> 1; if(B[mid] + Add[x] < w) ret = mid, l = mid + 1; else r = mid - 1; } return B[ret] + Add[x]; } inline int Sec_A(int x, int y, int w) { int ret = -1; if(bel[x] == bel[y]) { for(int i = x; i <= y; i ++) if(A[i] + Add[bel[x]] < w && A[i] + Add[bel[x]] > ret) ret = A[i] + Add[bel[x]]; return ret; } else { for(int i = x; i <= bel[x] * block; i ++) if(A[i] + Add[bel[x]] < w && A[i] + Add[bel[x]] > ret) ret = A[i] + Add[bel[x]]; for(int i = (bel[y] - 1) * block + 1; i <= y; i ++) if(A[i] + Add[bel[y]] < w && A[i] + Add[bel[y]] > ret) ret = A[i] + Add[bel[y]]; } for(int i = bel[x] + 1; i < bel[y]; i ++) { int imp = Calc(i, w); if(imp < w && imp > ret) ret = imp; } return ret; } int main() { n = read(); block = sqrt(n); for(int i = 1; i <= n; i ++) A[i] = read(); for(int i = 1; i <= n; i ++) bel[i] = (i - 1) / block + 1; if(n % block) cnt = n / block + 1; else cnt = n / block; for(int i = 1; i <= cnt; i ++) Work_sort(i); int T = n; while(T --) { int opt = read(), l = read(), r = read(), c = read(); if(!opt) Sec_G(l, r, c); else cout << Sec_A(l, r, c) << " "; } return 0; }
数列分块入门 4
区间加法，区间求和
没什么好说的
#include <iostream> #include <cstdio> #include <cmath> using namespace std; const int N = 5e4 + 10; #define gc getchar() inline int read() { int x = 0; char c = gc; while(c < '0' || c > '9') c = gc; while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = gc; return x; } #define LL long long LL A[N], Add[N], bel[N], W[N]; int n, block, cnt, Mod; void Sec_G(int x, int y, int w) { if(bel[x] == bel[y]) for(int i = x; i <= y; i ++) A[i] += w, W[bel[x]] += w; else { for(int i = x; i <= bel[x] * block; i ++) A[i] += w, W[bel[x]] += w; for(int i = (bel[y] - 1) * block + 1; i <= y; i ++) A[i] += w, W[bel[y]] += w; } for(int i = bel[x] + 1; i < bel[y]; i ++) Add[i] += w, W[i] += w * block; } inline int Sec_A(int x, int y) { LL ret = 0; if(bel[x] == bel[y]) for(int i = x; i <= y; i ++) ret += (A[i] + Add[bel[x]]) % Mod; else { for(int i = x; i <= bel[x] * block; i ++) ret += (A[i] + Add[bel[x]]) % Mod; for(int i = (bel[y] - 1) * block + 1; i <= y; i ++) ret += (A[i] + Add[bel[y]]) % Mod; } for(int i = bel[x] + 1; i < bel[y]; i ++) ret += W[i] % Mod; return ret% Mod; } int main() { n = read(); block = sqrt(n); for(int i = 1; i <= n; i ++) A[i] = read(); for(int i = 1; i <= n; i ++) bel[i] = (i - 1) / block + 1, W[bel[i]] += A[i]; if(n % block) cnt = n / block + 1; else cnt = n / block; int T = n; while(T --) { int opt = read(), l = read(), r = read(), c = read(); Mod = c + 1; if(!opt) Sec_G(l, r, c); else cout << Sec_A(l, r) << endl; } return 0; }
数列分块入门 5
区间开方，区间求和
一个数(合理)开几次根后就是0/1了
因此，只需记录每块的最大值，如果最大值是0/1就没必要开根
#include <iostream> #include <cstdio> #include <cmath> using namespace std; const int N = 5e4 + 10; #define gc getchar() inline int read() { int x = 0; char c = gc; while(c < '0' || c > '9') c = gc; while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = gc; return x; } #define LL long long LL A[N], Add[N], bel[N], W[N], Max[N]; int n, block, cnt, Mod; inline void Sec_G(int x, int y) { if(bel[x] == bel[y]) { if(!Max[bel[x]]) return ; if(Max[bel[x]] == 1) return ; for(int i = x; i <= y; i ++) { int C = A[i] - (int) sqrt(A[i]); W[bel[x]] -= C; A[i] = (int) sqrt(A[i]); } LL Max_A = 0; for(int i = (bel[x] - 1) * block + 1; i <= bel[x] * block; i ++) Max_A = max(Max_A, A[i]); Max[bel[x]] = Max_A; return ; } else { if(Max[bel[x]] && Max[bel[x]] != 1) { for(int i = x; i <= bel[x] * block; i ++) { int C = A[i] - (int) sqrt(A[i]); W[bel[x]] -= C; A[i] = (int) sqrt(A[i]); } LL Max_A = 0; for(int i = (bel[x] - 1) * block + 1; i <= bel[x] * block; i ++) Max_A = max(Max_A, A[i]); Max[bel[x]] = Max_A; } if(Max[bel[y]] && Max[bel[y]] != 1) { for(int i = (bel[y] - 1) * block + 1; i <= y; i ++) { int C = A[i] - (int) sqrt(A[i]); W[bel[y]] -= C; A[i] = (int) sqrt(A[i]); } LL Max_A = 0; for(int i = (bel[y] - 1) * block + 1; i <= bel[y] * block; i ++) Max_A = max(Max_A, A[i]); Max[bel[y]] = Max_A; } } for(int i = bel[x] + 1; i < bel[y]; i ++) { if(!Max[i] || Max[i] == 1) continue ; LL Max_A = 0; for(int j = (i - 1) * block + 1; j <= i * block; j ++) { int C = A[j] - (int) sqrt(A[j]); W[i] -= C; A[j] = (int) sqrt(A[j]); Max_A = max(Max_A, A[j]); } Max[i] = Max_A; } } inline int Sec_A(int x, int y) { LL ret = 0; if(bel[x] == bel[y] && Max[bel[x]]) for(int i = x; i <= y; i ++) ret += A[i]; else { for(int i = x; i <= bel[x] * block && Max[bel[x]]; i ++) ret += A[i]; for(int i = (bel[y] - 1) * block + 1; i <= y && Max[bel[y]]; i ++) ret += A[i]; } for(int i = bel[x] + 1; i < bel[y]; i ++) ret += W[i]; return ret; } int main() { n = read(); block = sqrt(n); for(int i = 1; i <= n; i ++) A[i] = read(); for(int i = 1; i <= n; i ++) bel[i] = (i - 1) / block + 1, W[bel[i]] += A[i], Max[bel[i]] = max(Max[bel[i]], A[i]); int T = n; while(T --) { int opt = read(), l = read(), r = read(), c = read(); if(!opt) Sec_G(l, r); else cout << Sec_A(l, r) << " "; } return 0; }
数列分块入门 6
单点插入，单点询问
数据随机，分块，对于每一块开动态数组，插入 + 查询比较容易实现
如果数据不随机，就有可能加到同一块中的数较多，影响效率
这样可以进行一定的插入操作之后重新分块
#include <iostream> #include <cstdio> #include <cmath> #include <vector> using namespace std; const int N = 1e5 + 10; int A[N << 1], n; vector <int> Vec[350]; int block, bel[N]; #define gc getchar() inline int read() { int x = 0; char c = gc; while(c < '0' || c > '9') c = gc; while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = gc; return x; } inline void Ins(int x, int a) { int now_size(0), Whi; for(int i = 1; ; i ++) { int Size = Vec[i].size(); now_size += Size; if(now_size >= x) { Whi = i; x -= (now_size - Size); break; } } Vec[Whi].insert(Vec[Whi].begin() + x - 1, a); } inline int Poi_A(int x) { int Whi_, now_size(0); for(int i = 1; ; i ++) { int Size = Vec[i].size(); now_size += Size; if(now_size >= x) { int iii = x - (now_size - Size); return Vec[i][iii - 1]; } } } int main() { n = read(); block = sqrt(n); for(int i = 1; i <= n; i ++) A[i] = read(); for(int i = 1; i <= n; i ++) bel[i] = (i - 1) / block + 1; for(int i = 1; i <= n; i ++) Vec[bel[i]].push_back(A[i]); int T = n; while(T --) { int opt = read(), l = read(), r = read(), c = read(); if(!opt) Ins(l, r); else cout << Poi_A(r) << endl; } return 0; }
数列分块入门 7
区间乘法，区间加法，单点询问
先乘后加，乘的时候相应的加法标记也要乘
#include <iostream> #include <cstdio> #include <cmath> #include <vector> using namespace std; const int N = 1e5 + 10; const int Mod = 1e4 + 7; #define LL long long LL A[N], Mul[N], Add[N], bel[N]; int n, cnt, block; #define gc getchar() inline int read() { int x = 0, f = 1; char c = gc; while(c < '0' || c > '9') {if(c == '-') f = -1; c = gc;} while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = gc; return x; } inline void Sec_Add(int x, int y, int w) { if(bel[x] == bel[y]) { for(int i = (bel[x] - 1) * block + 1; i <= bel[x] * block; i ++) A[i] = (A[i] * Mul[bel[x]] + Add[bel[x]]); Add[bel[x]] = 0; Mul[bel[x]] = 1; for(int i = x; i <= y; i ++) A[i] += w, A[i] %= Mod; return ; } else { for(int i = (bel[x] - 1) * block + 1; i <= bel[x] * block; i ++) A[i] = (A[i] * Mul[bel[i]] + Add[bel[i]]) % Mod; Add[bel[x]] = 0; Mul[bel[x]] = 1; for(int i = x; i <= bel[x] * block; i ++) A[i] += w, A[i] %= Mod; for(int i = (bel[y] - 1) * block + 1; i <= bel[y] * block; i ++) A[i] = (A[i] * Mul[bel[i]] + Add[bel[i]]) % Mod; Add[bel[y]] = 0; Mul[bel[y]] = 1; for(int i = (bel[y] - 1) * block + 1; i <= y; i ++) A[i] += w, A[i] %= Mod; } for(int i = bel[x] + 1; i < bel[y]; i ++) Add[i] += w, Add[i] %= Mod; } inline void Sec_Mul(int x, int y, int w) { if(bel[x] == bel[y]) { for(int i = (bel[x] - 1) * block + 1; i <= bel[x] * block; i ++) A[i] = (A[i] * Mul[bel[i]] + Add[bel[i]]) % Mod; Add[bel[x]] = 0; Mul[bel[x]] = 1; for(int i = x; i <= y; i ++) A[i] = (A[i] * w) % Mod; return ; } else { for(int i = (bel[x] - 1) * block + 1; i <= bel[x] * block; i ++) A[i] = (A[i] * Mul[bel[i]] + Add[bel[i]]) % Mod; Add[bel[x]] = 0; Mul[bel[x]] = 1; for(int i = x; i <= bel[x] * block; i ++) A[i] = (A[i] * w) % Mod; for(int i = (bel[y] - 1) * block + 1; i <= bel[y] * block; i ++) A[i] = (A[i] * Mul[bel[i]] + Add[bel[i]]) % Mod; Add[bel[y]] = 0; Mul[bel[y]] = 1; for(int i = (bel[y] - 1) * block + 1; i <= y; i ++) A[i] = (A[i] * w) % Mod; } for(int i = bel[x] + 1; i < bel[y]; i ++) Add[i] = (Add[i] * w) % Mod, Mul[i] = (Mul[i] * w) % Mod; } int main() { n = read(); for(int i = 1; i <= n; i ++) A[i] = read(); block = sqrt(n); for(int i = 1; i <= n; i ++) bel[i] = (i - 1) / block + 1, Mul[i] = 1; int T = n; while(T --) { int opt = read(), l = read(), r = read(), c = read(); if(opt == 0) Sec_Add(l, r, c); else if(opt == 1) Sec_Mul(l, r, c); else cout << (A[r] * Mul[bel[r]] + Add[bel[r]]) % Mod << " "; } return 0; }
数列分块入门 8
暴力
区间修改没有什么难度，这题难在区间查询比较奇怪，因为权值种类比较多，似乎没有什么好的维护方法。
模拟一些数据可以发现，询问后一整段都会被修改，几次询问后数列可能只剩下几段不同的区间了。
我们思考这样一个暴力，还是分块，维护每个分块是否只有一种权值，区间操作的时候，对于同权值的一个块就O(1)统计答案，否则暴力统计答案，并修改标记，不完整的块也暴力。
这样看似最差情况每次都会耗费O(n)的时间，但其实可以这样分析：
假设初始序列都是同一个值，那么查询是O(√n)，如果这时进行一个区间操作，它最多破坏首尾2个块的标记，所以只能使后面的询问至多多2个块的暴力时间，所以均摊每次操作复杂度还是O(√n)。
换句话说，要想让一个操作耗费O(n)的时间，要先花费√n个操作对数列进行修改。
初始序列不同值，经过类似分析后，就可以放心的暴力啦。
#include <iostream> #include <cstdio> #include <cmath> #include <cstdlib> #include <cstring> using namespace std; const int N = 1e5 + 10; int bel[N], A[N], bec[N]; int n; int block; #define gc getchar() inline int read() { int x = 0; char c = gc; while(c < '0' || c > '9') c = gc; while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = gc; return x; } int Sec_A(int x, int y, int c) { int ret(0); if(bel[x] == bel[y]) { if(bec[bel[x]] == c) ret = y - x + 1; else if(bec[bel[x]] == -1) for(int i = x; i <= y; i ++) if(A[i] == c) ret ++; if(~ bec[bel[x]]) for(int i = (bel[x] - 1) * block + 1; i < x; i ++) A[i] = bec[bel[x]]; if(~ bec[bel[x]]) for(int i = y + 1; i <= bel[x] * block; i ++) A[i] = bec[bel[x]]; for(int i = x; i <= y; i ++) A[i] = c; bec[bel[x]] = -1; } else { if(bec[bel[x]] == c) ret += bel[x] * block - x + 1; else if(bec[bel[x]] == -1) for(int i = x; i <= bel[x] * block; i ++) if(A[i] == c) ret ++; if(bec[bel[y]] == c) ret += y - ((bel[y] - 1) * block); else if(bec[bel[y]] == -1) for(int i = (bel[y] - 1) * block + 1; i <= y; i ++) if(A[i] == c) ret ++; for(int i = bel[x] + 1; i < bel[y]; i ++) { if(bec[i] == c) ret += block; else if(bec[i] == -1) for(int j = (i - 1) * block + 1; j <= i * block; j ++) if(A[j] == c) ret ++; } if(~ bec[bel[x]]) for(int i = (bel[x] - 1) * block + 1; i < x; i ++) A[i] = bec[bel[i]]; for(int i = x; i <= bel[x] * block; i ++) A[i] = c; bec[bel[x]] = -1; for(int i = (bel[y] - 1) * block + 1; i <= y; i ++) A[i] = c; if(~ bec[bel[y]]) for(int i = y + 1; i <= bel[y] * block; i ++) A[i] = bec[bel[i]]; bec[bel[y]] = -1; for(int i = bel[x] + 1; i < bel[y]; i ++) bec[i] = c; } return ret; } int main() { n = read(); for(int i = 1; i <= n; i ++) bec[i] = -1; for(int i = 1; i <= n; i ++) A[i] = read(); block = sqrt(n); for(int i = 1; i <= n; i ++) bel[i] = (i - 1) / block + 1; int T = n; while(T --) { int l = read(), r = read(), c = read(); cout << Sec_A(l, r, c) << " "; } return 0; }
数列分块入门 9
区间众数查询
陈立杰区间众数解题报告
#include<map> #include<set> #include<cmath> #include<stack> #include<queue> #include<cstdio> #include<vector> #include<cstring> #include<cstdlib> #include<iostream> #include<algorithm> #define mod 10007 #define pi acos(-1) #define inf 0x7fffffff #define ll long long using namespace std; ll read() { ll x=0,f=1; char ch=getchar(); while(ch<'0'||ch>'9') { if(ch=='-')f=-1; ch=getchar(); } while(ch>='0'&&ch<='9') { x=x*10+ch-'0'; ch=getchar(); } return x*f; } int n,blo,id; int v[50005],bl[50005]; int f[505][505]; map<int,int>mp; int val[50005],cnt[50005]; vector<int>ve[50005]; void pre(int x) { memset(cnt,0,sizeof(cnt)); int mx=0,ans=0; for(int i=(x-1)*blo+1; i<=n; i++) { cnt[v[i]]++; int t=bl[i]; if(cnt[v[i]]>mx||(cnt[v[i]]==mx&&val[v[i]]<val[ans])) ans=v[i],mx=cnt[v[i]]; f[x][t]=ans; } } int query(int l,int r,int x) { int t=upper_bound(ve[x].begin(),ve[x].end(),r)-lower_bound(ve[x].begin(),ve[x].end(),l); return t; } int query(int a,int b) { int ans,mx; ans=f[bl[a]+1][bl[b]-1]; mx=query(a,b,ans); for(int i=a; i<=min(bl[a]*blo,b); i++) { int t=query(a,b,v[i]); if(t>mx||(t==mx&&val[v[i]]<val[ans]))ans=v[i],mx=t; } if(bl[a]!=bl[b]) for(int i=(bl[b]-1)*blo+1; i<=b; i++) { int t=query(a,b,v[i]); if(t>mx||(t==mx&&val[v[i]]<val[ans]))ans=v[i],mx=t; } return ans; } int main() { n=read(); blo=200; for(int i=1; i<=n; i++) { v[i]=read(); if(!mp[v[i]]) { mp[v[i]]=++id; val[id]=v[i]; } v[i]=mp[v[i]]; ve[v[i]].push_back(i); } for(int i=1; i<=n; i++)bl[i]=(i-1)/blo+1; for(int i=1; i<=bl[n]; i++)pre(i); for(int i=1; i<=n; i++) { int a=read(),b=read(); if(a>b)swap(a,b); printf("%d ",val[query(a,b)]); } return 0; }
分块算法小结：
暴力算法
时间复杂度可以
空间允许
优美