[noi simulation] string matching (suffix automata Sam, Mo team, block)

Topic

1024MB,4s

Answer key

We build text strings into suffix automata , Then take the inquiry offline , Press the right endpoint to place it on the pattern string , Then you can find the pattern string where each prefix is SAM Corresponding node on , Get the number of occurrences of the substring , Update answers . This is $O(n^2)$ Of .

We can use the Mo team Algorithm , First preprocess each point on the suffix tree $p$ The largest of our ancestors $t\times len$ , And then because of The extended intervals in Mo's algorithm are all definite , We can put the interval in Mo team again “ offline ” To suffix tree , Calculate the corresponding SAM The points on the $p$ , Contribution is $p$ Times the current $p$ The effective length of , as well as $p$ The largest of our ancestors $t\times len$ . Time complexity $O(n\sqrt{n})$ .

But it's a little slow , We can actually prune directly with violence .

Go back to that $O(n^2)$ How to do it , We get the corresponding point of the current prefix $p$ after , stay $len[p]\sim len[fa[p]]-1$ Enumeration between the contributions , But in the process, if the answer is less than $p$ The largest of our ancestors $t\times len$ , You can just jump out of it . in addition , Don't jump one step at a time, father , You can jump directly to $t[q]\times len[q]$ maximal $p$ The ancestors of the $q$ . It can be proved that the upper limit of skipping ancestors is $O(\sqrt{n})$ Of , The number of violent enumeration is also very small , It is related to the size of the character set . The contribution of each enumeration is maintained in blocks ,$O(1)$ modify ,$O(\sqrt{n})$ Inquire about . This problem is limited to four seconds , This method only takes 500 milliseconds to round .

CODE

#include<map>
#include<set>
#include<cmath>
#include<ctime>
#include<queue>
#include<stack>
#include<random>
#include<bitset>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<unordered_map>
#pragma GCC optimize(2)
using namespace std;
#define MAXN 100005
#define LL long long
#define ULL unsigned long long
#define ENDL putchar('\n')
#define DB double
#define lowbit(x) (-(x) & (x))
#define FI first
#define SE second
#define PR pair<int,int>
#define UIN unsigned int
#define SQ 317
int xchar() {
    
	static const int maxn = 1000000;
	static char b[maxn];
	static int pos = 0,len = 0;
	if(pos == len) pos = 0,len = fread(b,1,maxn,stdin);
	if(pos == len) return -1;
	return b[pos ++];
}
// #define getchar() xchar()
inline LL read() {
    
	LL f = 1,x = 0;int s = getchar();
	while(s < '0' || s > '9') {
    if(s<0)return -1;if(s=='-')f=-f;s = getchar();}
	while(s >= '0' && s <= '9') {
    x = (x<<1) + (x<<3) + (s^48);s = getchar();}
	return f*x;
}
void putpos(LL x) {
    if(!x)return ;putpos(x/10);putchar((x%10)^48);}
inline void putnum(LL x) {
    
	if(!x) {
    putchar('0');return ;}
	if(x<0) putchar('-'),x = -x;
	return putpos(x);
}
inline void AIput(LL x,int c) {
    putnum(x);putchar(c);}

int n,m,s,o,k;
inline LL Max(LL a,LL b) {
    return a>b ? a:b;}
char s1[MAXN],s2[MAXN];
struct SAM{
    
	int s[26],len,fa;
	SAM(){
    memset(s,0,sizeof(s));len=fa=0;}
}sam[MAXN<<1];
int las = 1,cnt = 1;
int addsam(int c) {
    
	int p = las,np = (las = ++ cnt);
	sam[np].len = sam[p].len + 1;
	for(;p&&!sam[p].s[c];p=sam[p].fa)sam[p].s[c] = np;
	if(!p) sam[np].fa = 1;
	else {
    
		int q = sam[p].s[c];
		if(sam[q].len == sam[p].len + 1) sam[np].fa = q;
		else {
    
			int nq = ++ cnt; sam[nq] = sam[q];
			sam[nq].len = sam[p].len + 1;
			sam[np].fa = sam[q].fa = nq;
			for(;p&&sam[p].s[c]==q;p=sam[p].fa)sam[p].s[c]=nq;
		}
	}return np;
}
vector<int> g[MAXN<<1];
int ct[MAXN<<1],tp[MAXN<<1];
LL wt[MAXN<<1];
void dfs(int x) {
    for(int y:g[x])dfs(y),ct[x]+=ct[y];}
void dfs2(int x) {
    
	wt[x] = sam[x].len *1ll* ct[x];
	tp[x] = sam[x].fa;
	if(wt[tp[sam[x].fa]] >= wt[tp[x]]) tp[x] = tp[sam[x].fa];
	for(int y:g[x]) dfs2(y);
}
int bl[MAXN];
LL mw[MAXN],mb[MAXN];
inline void addp(int x,LL y) {
    
	mw[x] = Max(mw[x],y);
	mb[bl[x]] = Max(mb[bl[x]],y);
}
int ql[MAXN];
vector<int> bu[MAXN];
LL as[MAXN];
int main() {
    
	scanf("%s",s1 + 1);
	n = strlen(s1 + 1);
	for(int i = 1;i <= n;i ++) {
    
		bl[i] = i/SQ+1;
	}
	scanf("%s",s2 + 1);
	m = strlen(s2 + 1);
	for(int i = 1;i <= m;i ++) {
    
		s = addsam(s2[i]-'a');
		ct[s] ++;
	}
	for(int i = 2;i <= cnt;i ++) g[sam[i].fa].push_back(i);
	dfs(1); dfs2(1);
	int Q = read(),flag = 1;
	for(int i = 1;i <= Q;i ++) {
    
		ql[i] = read();o = read();
		bu[o].push_back(i);
		if(ql[i] > 1) flag = 0;
	}
	if(flag) {
    
		LL ans = 0;
		int p = 1,le = 0;
		for(int i = 1;i <= n;i ++) {
    
			int c = s1[i]-'a';
			while(p > 1 && !sam[p].s[c]) p = sam[p].fa,le = sam[p].len;
			if(sam[p].s[c]) p = sam[p].s[c],le ++;
			ans = max(ans,le*1ll*ct[p]);
			ans = max(ans,wt[tp[p]]);
			for(int x:bu[i]) {
    
				as[x] = ans;
			}
		}
		for(int i = 1;i <= Q;i ++) {
    
			AIput(as[i],'\n');
		} return 0;
	}
	int p = 1,le = 0;
	for(int i = 1;i <= n;i ++) {
    
		int c = s1[i]-'a';
		while(p > 1 && !sam[p].s[c]) p = sam[p].fa,le = sam[p].len;
		if(sam[p].s[c]) p = sam[p].s[c],le ++;
		int pp = p,ll = le;
		while(pp) {
    
			int nx = tp[pp];
			for(;ll > sam[sam[pp].fa].len && ll*1ll*ct[pp] > wt[nx];ll --) {
    
				addp(i-ll+1,ll*1ll*ct[pp]);
			} pp = nx; ll = sam[pp].len;
		}
		for(int x:bu[i]) {
    
			s = ql[x];
			for(int j = s;bl[j] == bl[s];j ++) as[x] = Max(as[x],mw[j]);
			for(int j = bl[s]+1;j <= bl[n];j ++) as[x] = Max(as[x],mb[j]);
		}
	}
	for(int i = 1;i <= Q;i ++) {
    
		AIput(as[i],'\n');
	}
	return 0;
}