hdu 4787 GRE Words Revenge 在线AC自动机

 

hdu 4787

GRE Words Revenge

Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 327680/327680 K (Java/Others)
Total Submission(s): 2505    Accepted Submission(s): 614

Problem Description

　　Now Coach Pang is preparing for the Graduate Record Examinations as George did in 2011. At each day, Coach Pang can:
　　"+w": learn a word w
　　"?p": read a paragraph p, and count the number of learnt words. Formally speaking, count the number of substrings of p which is a learnt words.
　　Given the records of N days, help Coach Pang to find the count. For convenience, the characters occured in the words and paragraphs are only '0' and '1'.

 

 

Input

　　The first line of the input file contains an integer T, which denotes the number of test cases. T test cases follow.
　　The first line of each test case contains an integer N (1 <= N <= 10⁵), which is the number of days. Each of the following N lines contains either "+w" or "?p". Both p and w are 01-string in this problem.
　　Note that the input file has been encrypted. For each string occured, let L be the result of last "?" operation. The string given to you has been shifted L times (the shifted version of string s₁s₂ ... s_k is s_ks₁s₂ ... s_k-1). You should decrypt the string to the original one before you process it. Note that L equals to 0 at the beginning of each test case.
　　The test data guarantees that for each test case, total length of the words does not exceed 10⁵ and total length of the paragraphs does not exceed 5 * 10⁶.

 

 

Output

　　For each test case, first output a line "Case #x:", where x is the case number (starting from 1).
　　And for each "?" operation, output a line containing the result.

 

 

Sample Input

2 3 +01 +01 ?01001 3 +01 ?010 ?011

 

 

Sample Output

Case #1: 2 Case #2: 1 0

 

 

Source

2013 Asia Chengdu Regional Contest

 

 

如果暴力每次插入后重建ac自动机,那么复杂度就是O(N*N),的,可以想到用分块(dalao是这么叫的);建立一个大的自动机,一个小的自动机,小的自动机规模是sqrt(N),大的是N,每次插入时在小的buf 里添加字串,重构自动机,当buf的规模超过sqrt(N)时,合并到大的自动机里大的重建,复杂度为O(sqrt(N) * N);合并就是buf 与ac 的字典树一起跑,如果ac里没有buf 的结点,就新建,然后权值|=buf.val[],(有节点也要| =,因为原来的不一定是单词,但现在是了);

 

 

//2017-07-15 19:28:18
//2017-07-15 20:10:10 
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<map>
#include"set"
#include"queue"
#include"vector"
#include"iomanip"
#include"cstring"
#define inf 1<<29
#define ll long long
#define re register
#define il inline
#define rep(i,a,b) for(register int i=a;i<=b;++i)
#define file(a) freopen(a".in","r",stdin),freopen(a".out","w",stdout);
using namespace std;
const int N=100010;
char tmp[5000100],s[5000100];
struct ACAutomation{
    int f[N],ch[N][2],lst[N],sz;
    int val[N];
    il void init() {
        sz=0;ch[0][0]=ch[0][1]=0;lst[0]=0;
        //val[0]=0;
    }
    il int newnode(){
        ++sz;
        ch[sz][0]=ch[sz][1]=f[sz]=lst[sz]=0;
        val[sz]=0;//
        return sz;
    }
    il bool search(char *s,int len) {
        re int t=0,c;
        for(re int i=0;i<len;++i){
            c=s[i]-'0';
            if(!ch[t][c])
                return 0;
            t=ch[t][c];
        }
        return val[t];
    }
    il void insert(char *s,int len){
        re int t=0,c;
        for(re int i=0;i<len;++i){
            c=s[i]-'0';
            if(!ch[t][c])
                ch[t][c]=newnode();
            t=ch[t][c];
        }
        val[t]=1;
    }
    il void build() {
        queue<int> q;
        for(re int i=0;i<2;++i){
            if(ch[0][i])
                q.push(ch[0][i]),f[ch[0][i]]=0,lst[ch[0][i]]=0;
        }
        re int r,v,u;
        while(!q.empty()) {
            r=q.front();q.pop();
            for(re int c=0;c<2;++c) {
                u=ch[r][c];
                if(!u) continue;
                q.push(u);
                v=f[r];
                while(v&&!ch[v][c]) v=f[v];
                f[u]=ch[v][c];
                lst[u] = val[f[u]] ? f[u] : lst[f[u]];
            }
        }
    }
    il int run(char *s,int len) {
        re int t=0,ans=0,c;
        for(re int i=0;i<len;++i) {
            c=s[i]-'0';
            while(t&&!ch[t][c]) t=f[t];
            t=ch[t][c];
            for(re int p=t;p;p=lst[p])
                ans+=val[p];
        }
        return ans;
    }
};
ACAutomation ac,buf;

inline int gi() {
    re int res=0,f=1;re char ch=getchar();
    while((ch<'0'||ch>'9')&&ch!='-') ch=getchar();
    if(ch=='-') f=-1,ch=getchar();
    while(ch>='0'&&ch<='9') res=res*10+ch-'0',ch=getchar();
    return res*f;
}
il void bfs() {
    queue<int> U,V;
    U.push(0),V.push(0);
    re int u,v,r1,r2;
    while(!U.empty()) {
        u=U.front(),v=V.front();
        U.pop(),V.pop();
        for(re int c=0;c<2;++c)
            if(buf.ch[u][c]) {
                r1=buf.ch[u][c],r2=ac.ch[v][c];
                if(!r2) {
                    ac.ch[v][c]=ac.newnode();
                }
                r2=ac.ch[v][c];
                ac.val[r2]|=buf.val[r1];
                U.push(r1),V.push(r2);
            }
    }
}
il void go() {
    bfs();
    buf.init();
    ac.build();    
}
int main(){
    file("Y");
    re int cas=gi();
    rep(tt,1,cas) {
        printf("Case #%d:\n",tt);
        ac.init();
        buf.init();
        re int n=gi();
        re int l=0;
        rep(qyp,1,n) {
            scanf("%s",tmp);
            re int len=strlen(tmp);
            re int p=l%(len-1);
            s[0]=tmp[0];
            for(re int i=1;i<=p;i++) s[i+len-1-p]=tmp[i];
            for(re int i=1;i<len-p;++i) s[i]=tmp[i+p];
            if(s[0]=='+') {
                if(buf.search(s+1,len-1)||ac.search(s+1,len-1)) continue;
                buf.insert(s+1,len-1);
                buf.build();
                if(buf.sz>333) go();
            }
            else {
                l=ac.run(s+1,len-1) + buf.run(s+1,len-1);
                printf("%d\n",l);
            }
        }
    }
    return 0;
}