HDU 4787 GRE Words Revenge

分块思路：

因为模式串和母串交叉给出，正常来说应该是，每次询问前都要getFail，这样显然会超时）

所以我们用一个小型ac自动机 buf ， 每次插入都插入到 buf 中，并重建一下buf 的getFail

若buf的节点数 ＞ 2000，则把其中节点添加到 ac自动机上

时间复杂度为 O(n*sqrt(n)) 

GRE Words Revenge

Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 327680/327680 K (Java/Others)
Total Submission(s): 1004    Accepted Submission(s): 215

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 <= 10 ⁵), 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 s ₁s ₂ ... s _k is s _ks ₁s ₂ ... 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 10 ⁵ and total length of the paragraphs does not exceed 5 * 10 ⁶.

 

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

 

#pragma comment( linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<vector>
#include<cstring>
#include<queue>
using namespace std;
#define prt(k) cout<<#k" = "<<k<<endl;
const int M = 500010;
struct AC
{
    int ch[M][2]; int t[M][2];
    int L, root;
    int fail[M];
    int end[M];

    void init()
    {
        L=1;
        memset(ch,0,sizeof ch);
        memset(end,0,sizeof end);
        memset(fail,0,sizeof fail);
        root=0;
    }
    AC() { init(); }
    int id(char a) { return a-'0'; }
    bool search(char s[])
    {
        int u=0;
        for(int i=0;s[i];i++) {
            u = ch[u][s[i]-'0'];
            if(u==0) return false;
        }
        return end[u];
    }
    void insert(char s[])
    {
        int u = 0;
        for(int i=0;s[i];i++) {
            int c = id(s[i]);
            if(!ch[u][c]) ch[u][c]=L++;
            u = ch[u][c];
        }
        end[u]=1;
    }
    void BUILD()
    {
        queue<int> q;
        for(int i=0;i<2;i++)
            if(ch[0][i])
            q.push(ch[0][i]);

        while(!q.empty())
        {
            int r=q.front(); q.pop();
            for(int c=0;c<2;c++) {
                int u = ch[r][c];
                if(u==0) continue;
                q.push(u);
                int v = fail[r];
                while(v && ch[v][c]==0) v=fail[v];
                fail[u] = ch[v][c];
            }
        }

    }
    int query(char s[])
    {
        int j = 0, ans = 0;
        for(int i = 0; s[i] ; i++){
            int c = s[i]-'0';
            while(j && ch[j][c]==0) j = fail[j];
            j = ch[j][c];

            int temp = j;
            while(temp){
                ans += end[temp];
                temp = fail[temp];
            }
        }
        return ans;
    }
};
AC ac, buf;
const int N = 10*M;
int n, Q;
char s[N], tmp[N];
void dfs(int u, int v){
    for(int i = 0;i < 2;i++){
        if( buf.ch[v][i] )
        {
            int e2 = buf.ch[v][i];
            if(! ac.ch[u][i])
            {
                memset(ac.ch[ac.L], 0, sizeof(ac.ch[ac.L]));
                ac.ch[u][i] = ac.L++;
            }
            int e1 = ac.ch[u][i];
            ac.end[e1] |= buf.end[e2];
            dfs(e1, e2);
        }
    }
}
void join()
{
    dfs(0,0);
    buf.init();
    ac.BUILD();
}
int main()
{
    int re; cin>>re; int ca=1;
    while(re--) {
        cin>>Q;
        printf("Case #%d:\n", ca++);
        ac.init();
        buf.init();
        int L = 0;
        while(Q--)
        {
            scanf("%s",tmp);
            int len = strlen(tmp+1);
            s[0] = tmp[0];
            for(int i=0;i<len;i++)
                s[i+1] = tmp[1+(i+L+len)%len];
            s[len+1]=0;

            if(s[0]=='+') {
                if(buf.search(s+1) || ac.search(s+1)) continue;
                buf.insert(s+1);
                buf.BUILD();
                if(buf.L>2000) join();
            }
            else {
                L = buf.query(s+1) + ac.query(s+1);
                printf("%d\n", L);
            }
        }
    }
    return 0;
}