題目鏈接： BZOJ - 3236? ?BZOJ - 3809

?

算法一：莫隊

首先，單純的莫隊算法是很好想的，就是用普通的第一關(guān)鍵字為 l 所在塊，第二關(guān)鍵字為 r 的莫隊。

這樣每次端點移動添加或刪除一個數(shù)字，用樹狀數(shù)組維護所求的信息就是很容易的。由于這里有 logn復(fù)雜度，所以這樣移動端點的復(fù)雜度還是挺高的。

于是 BZOJ-3236 的時限 100s，我的代碼跑了 98s，險過......

Paste一個BZOJ-3236的純莫隊代碼：

#include <iostream> #include <cstdlib> #include <cstdio> #include <cmath> #include <algorithm> #include <cstring>using namespace std;inline void Read(int &Num) {char c; c = getchar();while (c < '0' || c > '9') c = getchar();Num = c - '0'; c = getchar();while (c >= '0' && c <= '9') {Num = Num * 10 + c - '0';c = getchar();} }const int MaxN = 100000 + 5, MaxM = 1000000 + 5;int n, m, BlkSize; int A[MaxN], Cnt[MaxN], T1[MaxN], T2[MaxN];struct Query {int l, r, a, b, Pos, e, Ans1, Ans2;Query() {}Query(int x, int y, int p, int q, int o) {l = x; r = y; a = p; b = q; Pos = o;}bool operator < (const Query &q) const {if (e == q.e) return r < q.r;return e < q.e;} } Q[MaxM];inline bool Cmp(Query q1, Query q2) {return q1.Pos < q2.Pos; }inline void Add1(int x, int Num) {for (int i = x; i <= n; i += i & -i) T1[i] += Num; } inline int Get1(int x) {if (x == 0) return 0; //Notice!int ret = 0;for (int i = x; i; i -= i & -i) ret += T1[i];return ret; }inline void Add2(int x, int Num) {for (int i = x; i <= n; i += i & -i) T2[i] += Num; } inline int Get2(int x) {if (x == 0) return 0; //Notice!int ret = 0;for (int i = x; i; i -= i & -i) ret += T2[i];return ret; }inline void Add_Num(int x) {if (Cnt[x] == 0) Add2(x, 1);++Cnt[x];Add1(x, 1); } inline void Del_Num(int x) {--Cnt[x];Add1(x, -1);if (Cnt[x] == 0) Add2(x, -1); }void Pull(int f, int x, int y) {if (x == y) return;if (f == 0) if (x < y) for (int i = x; i < y; ++i) Del_Num(A[i]);else for (int i = x - 1; i >= y; --i) Add_Num(A[i]);else if (x < y) for (int i = x + 1; i <= y; ++i) Add_Num(A[i]);else for (int i = x; i > y; --i) Del_Num(A[i]); }int main() {Read(n); Read(m);BlkSize = (int)sqrt((double)n);for (int i = 1; i <= n; ++i) Read(A[i]);int l, r, a, b;for (int i = 1; i <= m; ++i) {Read(l); Read(r); Read(a); Read(b);Q[i] = Query(l, r, a, b, i);Q[i].e = (l - 1) / BlkSize + 1;}sort(Q + 1, Q + m + 1);memset(Cnt, 0, sizeof(Cnt));memset(T1, 0, sizeof(T1));memset(T2, 0, sizeof(T2));for (int i = Q[1].l; i <= Q[1].r; ++i) Add_Num(A[i]);Q[1].Ans1 = Get1(Q[1].b) - Get1(Q[1].a - 1);Q[1].Ans2 = Get2(Q[1].b) - Get2(Q[1].a - 1);for (int i = 2; i <= m; ++i) {if (Q[i].r < Q[i - 1].l) {Pull(0, Q[i - 1].l, Q[i].l);Pull(1, Q[i - 1].r, Q[i].r);}else {Pull(1, Q[i - 1].r, Q[i].r);Pull(0, Q[i - 1].l, Q[i].l);}Q[i].Ans1 = Get1(Q[i].b) - Get1(Q[i].a - 1);Q[i].Ans2 = Get2(Q[i].b) - Get2(Q[i].a - 1);}sort(Q + 1, Q + m + 1, Cmp);for (int i = 1; i <= m; ++i) printf("%d %d\n", Q[i].Ans1, Q[i].Ans2);return 0; }

?

算法二：莫隊+分塊

這是一個神奇的做法，還是用莫隊轉(zhuǎn)移區(qū)間端點，但是每次移動端點都是O(1)的，不再用樹狀數(shù)組，而是將 [1,n] 的數(shù)值分成大小為 sqrt(n) 的塊。

莫隊時加入一個數(shù)，如果它之前不存在，就將它所在的塊的數(shù)組值加一，刪除時類似。

然后每個詢問轉(zhuǎn)移完區(qū)間之后用分塊的方法查詢答案，中間的整塊直接查，兩邊的零散的數(shù)就暴力枚舉，這樣每個詢問就是 sqrt(n) 的。

總復(fù)雜度也大約是 O(n^1.5) ，主要就是把移動區(qū)間端點變?yōu)榱?O(1)，十分神奇！

我用這個算法寫了 3809，代碼如下：

#include <iostream> #include <cstdio> #include <cstdlib> #include <cmath> #include <algorithm> #include <cstring>using namespace std;inline void Read(int &Num) {char c; c = getchar();while (c < '0' || c > '9') c = getchar();Num = c - '0'; c = getchar();while (c >= '0' && c <= '9') {Num = Num * 10 + c - '0';c = getchar();} }const int MaxN = 100000 + 5, MaxM = 1000000 + 5, MaxBlk = 350 + 5;int n, m, BlkSize, TotBlk; int A[MaxN], Cnt[MaxN], T[MaxBlk], L[MaxBlk], R[MaxBlk], Ans[MaxM], Blk[MaxN];struct Query {int l, r, Index, a, b;Query() {}Query(int f, int x, int y, int p, int q) {Index = f; l = x; r = y; a = p; b = q;}bool operator < (const Query &b) const {if (Blk[l] == Blk[b.l]) return r < b.r;return Blk[l] < Blk[b.l];} } Q[MaxM];inline void Add_Num(int x) {if (Cnt[x] == 0) ++T[Blk[x]];++Cnt[x]; }inline void Del_Num(int x) {--Cnt[x];if (Cnt[x] == 0) --T[Blk[x]]; }void Pull(int f, int x, int y) {if (x == y) return;if (f == 0) { if (x < y) for (int i = x; i < y; ++i) Del_Num(A[i]);else for (int i = x - 1; i >= y; --i) Add_Num(A[i]);}else {if (x < y) for (int i = x + 1; i <= y; ++i) Add_Num(A[i]);else for (int i = x; i > y; --i) Del_Num(A[i]);} }int GetAns(int a, int b) {int x, y, ret;x = Blk[a]; if (L[x] != a) ++x;y = Blk[b]; if (R[y] != b) --y;ret = 0;if (x > y) {for (int i = a; i <= b; ++i) if (Cnt[i] > 0) ++ret;}else {for (int i = x; i <= y; ++i) ret += T[i];for (int i = a; i < L[x]; ++i) if (Cnt[i] > 0) ++ret;for (int i = b; i > R[y]; --i) if (Cnt[i] > 0) ++ret; }return ret; }int main() {Read(n); Read(m);for (int i = 1; i <= n; ++i) Read(A[i]);BlkSize = (int)sqrt((double)n);TotBlk = (n - 1) / BlkSize + 1;for (int i = 1; i <= TotBlk; ++i) {L[i] = (i - 1) * BlkSize + 1;R[i] = i * BlkSize;}R[TotBlk] = n;for (int i = 1; i <= n; ++i) Blk[i] = (i - 1) / BlkSize + 1;int l, r, a, b;for (int i = 1; i <= m; ++i) {Read(l); Read(r); Read(a); Read(b);Q[i] = Query(i, l, r, a, b);}sort(Q + 1, Q + m + 1);memset(Cnt, 0, sizeof(Cnt));memset(T, 0, sizeof(T));for (int i = Q[1].l; i <= Q[1].r; ++i) Add_Num(A[i]);Ans[Q[1].Index] = GetAns(Q[1].a, Q[1].b);for (int i = 2; i <= m; ++i) {if (Q[i].r < Q[i - 1].l) {Pull(0, Q[i - 1].l, Q[i].l);Pull(1, Q[i - 1].r, Q[i].r);}else {Pull(1, Q[i - 1].r, Q[i].r);Pull(0, Q[i - 1].l, Q[i].l);}Ans[Q[i].Index] = GetAns(Q[i].a, Q[i].b);}for (int i = 1; i <= m; ++i) printf("%d\n", Ans[i]);return 0; }

　　

轉(zhuǎn)載于:https://www.cnblogs.com/JoeFan/p/4246291.html

總結(jié)

以上是生活随笔為你收集整理的[BZOJ 3236] [Ahoi2013] 作业 [BZOJ 3809] 【莫队（+分块）】的全部內(nèi)容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網(wǎng)站內(nèi)容還不錯，歡迎將生活随笔推薦給好友。