[BZOJ 3236] [Ahoi2013] 作业 && [BZOJ 3809] 【莫队(+分块)】

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;
}

  

posted @ 2015-01-24 18:53  JoeFan  阅读(338)  评论(0编辑  收藏  举报