题目链接： BZOJ - 3236 BZOJ - 3809
算法一：莫队
首先，单纯的莫队算法是很好想的，就是用普通的第一关键字为 l 所在块，第二关键字为 r 的莫队。
这样每次端点移动添加或删除一个数字，用树状数组维护所求的信息就是很容易的。由于这里有 logn复杂度，所以这样移动端点的复杂度还是挺高的。
于是 BZOJ-3236 的时限 100s，我的代码跑了 98s，险过......
Paste一个BZOJ-3236的纯莫队代码：
#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：莫队+分块
这是一个神奇的做法，还是用莫队转移区间端点，但是每次移动端点都是O(1)的，不再用树状数组，而是将 [1,n] 的数值分成大小为 sqrt(n) 的块。
莫队时加入一个数，如果它之前不存在，就将它所在的块的数组值加一，删除时类似。
然后每个询问转移完区间之后用分块的方法查询答案，中间的整块直接查，两边的零散的数就暴力枚举，这样每个询问就是 sqrt(n) 的。
总复杂度也大约是 O(n^1.5) ，主要就是把移动区间端点变为了 O(1)，十分神奇！
我用这个算法写了 3809，代码如下：
#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;
}
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