CodeForces 547E Mike and Friends [后缀树组+主席树]

最新推荐文章于 2022-09-02 13:09:20 发布

GrassTreeFlower

最新推荐文章于 2022-09-02 13:09:20 发布

阅读量1.4k

点赞数

CC 4.0 BY-SA版权

分类专栏：数据结构--主席树字符串--后缀数组/树/自动机文章标签： codeforces suffix 后缀树组主席树可持久化数据结构

本文链接：https://blog.youkuaiyun.com/GrassTreeFlower/article/details/50756749

数据结构--主席树同时被 2 个专栏收录

4 篇文章

订阅专栏

字符串--后缀数组/树/自动机

2 篇文章

订阅专栏

本文介绍了一种利用AC自动机建立失败树和后缀数组来解决子串查询问题的高效算法。通过AC自动机实现字符串匹配，结合后缀数组快速查找区间内所有子串的出现次数，从而实现复杂度优化。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Description

What-The-Fatherland is a strange country! All phone numbers there are strings consisting of lowercase English letters. What is double strange that a phone number can be associated with several bears!

In that country there is a rock band called CF consisting of n bears (including Mike) numbered from 1 to n.

$\text{[math]}$

Phone number of i-th member of CF is s_i. May 17th is a holiday named Phone Calls day. In the last Phone Calls day, everyone called all the numbers that are substrings of his/her number (one may call some number several times). In particular, everyone called himself (that was really strange country).

Denote as call(i, j) the number of times that i-th member of CF called the j-th member of CF.

The geek Mike has q questions that he wants to ask you. In each question he gives you numbers l, r and k and you should tell him the number

$\text{[math]}$

题意：给出N个串，M个询问，对于每个询问L,R,K，给出串L到串R的所有串中串K作为子串出现了几次。

解法：正解是AC自动机建Fail树然后根据DFS序乱搞，但是脑洞太小，只能想出比较朴素的后缀数组+主席树的解法。

可知，将所有串并成一个串做完SA后，是可以二分求出每个串作为子串的区间（Lx , Rx）的 (SA中的下标进行二分)，二分过程只要比较一下LCP与原串的长度即可。然后只要询问这个区间中所有的后缀所属于的串的编号，在询问的区间（L,R）中的个数。这个过程可以用主席树实现。比较卡常数，不过稍微改下就能过。

代码：（这里后缀数组使用了DC3实现）

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
#include<iostream>
#include<stdlib.h>
#include<set>
#include<map>
#include<queue>
#include<vector>
#include<bitset>
#pragma comment(linker, "/STACK:1024000000,1024000000")
template <class T>
bool scanff(T &ret){ //Faster Input
    char c; int sgn; T bit=0.1;
    if(c=getchar(),c==EOF) return 0;
    while(c!='-'&&c!='.'&&(c<'0'||c>'9')) c=getchar();
    sgn=(c=='-')?-1:1;
    ret=(c=='-')?0:(c-'0');
    while(c=getchar(),c>='0'&&c<='9') ret=ret*10+(c-'0');
    if(c==' '||c=='\n'){ ret*=sgn; return 1; }
    while(c=getchar(),c>='0'&&c<='9') ret+=(c-'0')*bit,bit/=10;
    ret*=sgn;
    return 1;
}
#define inf 1073741823
#define llinf 4611686018427387903LL
#define PI acos(-1.0)
#define lth (th<<1)
#define rth (th<<1|1)
#define rep(i,a,b) for(int i=int(a);i<=int(b);i++)
#define drep(i,a,b) for(int i=int(a);i>=int(b);i--)
#define gson(i,root) for(int i=ptx[root];~i;i=ed[i].next)
#define tdata int testnum;scanff(testnum);for(int cas=1;cas<=testnum;cas++)
#define mem(x,val) memset(x,val,sizeof(x))
#define mkp(a,b) make_pair(a,b)
#define findx(x) lower_bound(b+1,b+1+bn,x)-b
#define pb(x) push_back(x)
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
#define lowbit(x) (x&(-x))

#define NN 800040
#define SIZZ 256
#define F(x) ((x)/3+((x)%3==1?0:tb))
#define G(x) ((x)<tb?(x)*3+1:((x)-tb)*3+2)

int aa[3*NN],bb[NN],cc[NN],xx[NN],yy[NN];
int sa[3*NN],height[NN],rank[NN];
int hf[NN][25];
char s[NN];

int c0(int *r,int a,int b){
	return r[a]==r[b]&&r[a+1]==r[b+1]&&r[a+2]==r[b+2];
}
int c12(int k,int *r,int a,int b){
	if(k==2)return r[a]<r[b]||r[a]==r[b]&&c12(1,r,a+1,b+1);
	else return r[a]<r[b]||r[a]==r[b]&&cc[a+1]<cc[b+1];
}

void sort(int *r,int *a,int *b,int n,int m){
	int i;
	for(i=0;i<n;i++)cc[i]=r[a[i]];
	for(i=0;i<m;i++)bb[i]=0;
	for(i=0;i<n;i++)bb[cc[i]]++;
	for(i=1;i<m;i++)bb[i]+=bb[i-1];
	for(i=n-1;i>=0;i--)b[--bb[cc[i]]]=a[i];
}
void dc3(int *r,int *sa,int n,int m){
	int i,j,ta=0,tb=(n+1)/3,tc=0,p;
	int *rn=r+n,*sn=sa+n;
	r[n]=r[n+1]=0;
	for(i=0;i<n;i++)if(i%3!=0)xx[tc++]=i;
	sort(r+2,xx,yy,tc,m);
	sort(r+1,yy,xx,tc,m);
	sort(r,xx,yy,tc,m);
	for(p=1,rn[F(yy[0])]=0,i=1;i<tc;i++)
	rn[F(yy[i])]=c0(r,yy[i-1],yy[i])?p-1:p++;
	if(p<tc)dc3(rn,sn,tc,p);
	else for(i=0;i<tc;i++)sn[rn[i]]=i;
	for(i=0;i<tc;i++)if(sn[i]<tb)yy[ta++]=sn[i]*3;
	if(n%3==1)yy[ta++]=n-1;
	sort(r,yy,xx,ta,m);
	for(i=0;i<tc;i++)cc[yy[i]=G(sn[i])]=i;
	for(i=0,j=0,p=0;i<ta&&j<tc;p++)
	sa[p]=c12(yy[j]%3,r,xx[i],yy[j])?xx[i++]:yy[j++];
	for(;i<ta;p++)sa[p]=xx[i++];
	for(;j<tc;p++)sa[p]=yy[j++];
}
void calh(int n){
	int i,j,k=0;
	for(i=0;i<n;i++)rank[sa[i]]=i;
	for(i=0;i<n;i++){
		if(k)k--;
		j=sa[rank[i]-1];
		if(rank[i]-1<0)continue;
		while(i+k<n-1&&j+k<n-1&&aa[i+k]==aa[j+k])k++;
		height[rank[i]]=k;
	}
	for(i=0;i<n;i++)hf[i][0]=height[i];
	for(j=1;j<=log(n*1.0)/log(2.0);j++){
		for(i=0;i<n;i++){
			if(i+(1<<j)<n){
				hf[i][j]=min(hf[i][j-1],hf[i+(1<<(j-1))][j-1]);
			}
		}
	}
}

int lcp(int x,int y){
	int i=rank[x];
	int j=rank[y];
	if(j<i)swap(i,j);
	int k=log(j-i)*1.0/log(2.0);
	return min(hf[i+1][k],hf[j-(1<<k)+1][k]);
}


int n,m,qn,ql,qr;
int len[NN];
int idx[NN];
int belong[NN];


#define MM 10000000
int tot=0;
int t[NN];
int c[MM],lson[MM],rson[MM];

int build(int l,int r){
    int x=++tot;
    if(l!=r){
        int m=(l+r)>>1;
        lson[x]=build(l,m);
        rson[x]=build(m+1,r);
    }
    return x;
}


void update(int l,int r,int pos,int pre,int cur,int val){
    lson[cur]=lson[pre];
    rson[cur]=rson[pre];
    c[cur]=c[pre]+val;
    if(l==r)return;
    int m=(l+r)>>1;
    if(pos<=m)update(l,m,pos,lson[pre],lson[cur]=++tot,val);
    else update(m+1,r,pos,rson[pre],rson[cur]=++tot,val);
}

int gettop(int idx,int len){
    int l=0,r=idx;
    while(l+1<r){
        int m=(l+r)>>1;
        if(lcp(sa[m],sa[idx])>=len)r=m;
        else l=m;
    }
    if(lcp(sa[l],sa[idx])>=len)return l;
    return r;
}

int getbutton(int idx,int len){
    int l=idx,r=n-1;
    while(l+1<r){
        int m=(l+r)>>1;
        if(lcp(sa[m],sa[idx])>=len)l=m;
        else r=m;
    }
    if(lcp(sa[r],sa[idx])>=len)return r;
    return l;
}

int query(int l,int r,int x){
    if(l>=ql&&r<=qr)return c[x];
    int sum=0;
    int m=(l+r)>>1;
    if(m>=ql)sum+=query(l,m,lson[x]);
    if(m<qr)sum+=query(m+1,r,rson[x]);
    return sum;
}

int lx[NN];
int rx[NN];
main(){
    n=0;
    scanff(m);
    scanff(qn);
	rep(i,1,m){
		scanf("%s",s);
		len[i]=strlen(s);
		idx[i]=n;
		rep(j,0,len[i]-1){
		    aa[n++]=s[j];
		    belong[n-1]=i;
		}
		aa[n++]=SIZZ+i;
		belong[n-1]=i;
	}
	dc3(aa,sa,n,SIZZ+m+10);
	calh(n);
	build(1,m);
	rep(i,0,n)t[i]=1;
    rep(i,1,n){
        if(aa[sa[i-1]]>=SIZZ){
            n=i;
            break;
        }
        t[i]=++tot;
        update(1,m,belong[sa[i-1]],t[i-1],t[i],1);
    }
    rep(i,1,qn){
        int k;
        scanff(ql);
        scanff(qr);
        scanff(k);
        int l,r;
        if(lx[k])l=lx[k];
        else l=gettop(rank[idx[k]],len[k])+1,lx[k]=l;
        if(rx[k])r=rx[k];
        else r=getbutton(rank[idx[k]],len[k])+1,rx[k]=r;
        printf("%d\n",query(1,m,t[r])-query(1,m,t[l-1]));
    }
}