[BZOJ 3207] 花神的嘲讽计划Ⅰ【Hash + 可持久化线段树】

题目链接：BZOJ - 3207

题目分析

先使用Hash，把每个长度为 k 的序列转为一个整数，然后题目就转化为了询问某个区间内有没有整数 x 。

这一步可以使用可持久化线段树来做，虽然感觉可以有更简单的做法，但是我没有什么想法...

代码

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cstring>
#include <cmath>

using namespace std;

const int MaxN = 200000 + 5, P = 233, Mod = 3371723, MaxNode = 8000000 + 5; 

int n, m, k, en, TL_Index, Node_Index;
int A[MaxN], B[MaxN], Root[MaxN], Lc[MaxNode], Rc[MaxNode], T[MaxNode];

struct HashNode 
{
	int Pos, TL;
	HashNode *Next;
} H[MaxN], *Ph = H, *Hash[Mod + 5];

bool Cmp(int *AA, int x, int *AB, int y) {
	for (int i = 0; i < k; ++i) 
		if (AA[x + i] != AB[y + i]) return false;
	return true;
}

void Insert(int &Now, int Last, int s, int t, int x) {
	if (Now == 0) Now = ++Node_Index;
	if (s == t) {
		T[Now] = T[Last] + 1;
		return;
	}
	int m = (s + t) >> 1;
	if (x <= m) {
		Rc[Now] = Rc[Last];
		Insert(Lc[Now], Lc[Last], s, m, x);
	}
	else {
		Lc[Now] = Lc[Last];
		Insert(Rc[Now], Rc[Last], m + 1, t, x);
	}
}

int main() 
{
	scanf("%d%d%d", &n, &m, &k);
	for (int i = 1; i <= n; ++i) scanf("%d", &A[i]);
	en = n - k + 1;
	int HN, TL_i;
	HashNode *Now;
	TL_Index = 0;
	Node_Index = 0;
	for (int i = 1; i <= en; ++i) { 
		HN = 0;
		for (int j = i; j < i + k; ++j) {
			HN = HN * P + A[j];
			if (HN > Mod) HN %= Mod;
		}
		Now = Hash[HN];
		TL_i = 0;
		while (Now != NULL) {
			if (Cmp(A, i, A, Now -> Pos)) {
				TL_i = Now -> TL;
				break;
			}
			Now = Now -> Next;
		}
		if (TL_i == 0) {
			++Ph; Ph -> Pos = i;
			Ph -> TL = TL_i = ++TL_Index;
			Ph -> Next = Hash[HN]; Hash[HN] = Ph;
		}
		Insert(Root[i], Root[i - 1], 1, n, TL_i);
	}
	int l, r, s, t, mid, x, y;
	for (int i = 1; i <= m; ++i) {
		scanf("%d%d", &l, &r);
		for (int j = 1; j <= k; ++j) scanf("%d", &B[j]);
		HN = 0;
		for (int j = 1; j <= k; ++j) {
			HN = HN * P + B[j];
			if (HN > Mod) HN %= Mod;
		}
		TL_i = 0;
		Now = Hash[HN];
		while (Now != NULL) {
			if (Cmp(B, 1, A, Now -> Pos)) {
				TL_i = Now -> TL;
				break;
			}
			Now = Now -> Next;
		}
		if (TL_i == 0 || r - l + 1 < k) printf("Yes
");
		else {
			r = r - k + 1;
			x = Root[l - 1]; y = Root[r];
			s = 1; t = n;
			while (s != t) {
				mid = (s + t) >> 1;
				if (TL_i <= mid) {
					x = Lc[x]; y = Lc[y];
					t = mid;
				}
				else {
					x = Rc[x]; y = Rc[y];
					s = mid + 1;
				}
			}
			if (T[y] - T[x] > 0) printf("No
");
			else printf("Yes
");
		}
	}
	return 0;
}