题目链接: BZOJ - 3236 BZOJ - 3809
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......
However..BZOJ-3809 的出题人(SLYZ的神犇)就没有这么善良了!直接内存限制 28MB 就直接把我的莫队卡成 MLE!这是处心积虑卡莫队的恶劣行为!严正抗议!
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:
[BZOJ 3236] [Ahoi2013] 作业 && [BZOJ 3809] 【莫队 | 分块】
原文:http://www.cnblogs.com/JoeFan/p/4246291.html