BZOJ 2555 SubString 后缀自动机

题目大意：

如题面所述, 在线状态下接受添加字符和查询某个字符串出现的次数

加上了一些加密的条件使得被迫在线查询

大致思路：

就是后缀自动机的裸题吧=_=

没什么特别的, 直接贴代码好了, 不懂的看代码细节吧

代码如下：

Result  :  Accepted     Memory  :  140140 KB     Time  :  6044 ms

/*
 * Author: Gatevin
 * Created Time:  2015/5/5 17:57:21
 * File Name: Rin_Tohsaka.cpp
 */
#include<iostream>
#include<sstream>
#include<fstream>
#include<vector>
#include<list>
#include<deque>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<bitset>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cctype>
#include<cmath>
#include<ctime>
#include<iomanip>
using namespace std;
const double eps(1e-8);
typedef long long lint;
#define foreach(e, x) for(__typeof(x.begin()) e = x.begin(); e != x.end(); ++e)
#define SHOW_MEMORY(x) cout<<sizeof(x)/(1024*1024.)<<"MB"<<endl

#define maxm 600010
#define maxn 1200010
#define maxl 3000010

struct Suffix_Automation
{
    struct State
    {
        State *par;
        State *go[26];
        int right, val;
        void init(int _val = 0)
        {
            val = _val, par = 0, right = 0;
            memset(go, 0, sizeof(go));
        }
    };
    State *root, *last, *cur;
    State nodePool[maxn];
    State* newState(int val = 0)
    {
        cur->init(val);
        return cur++;
    }
    void init()
    {
        cur = nodePool;
        root = newState();
        last = root;
    }
    void extend(int w)
    {
        State *p = last;
        State *np = newState(p->val + 1);
        np->right = 1;
        while(p && p->go[w] == 0)
        {
            p->go[w] = np;
            p = p->par;
        }
        if(p == 0)
        {
            np->par = root;
        }
        else
        {
            State *q = p->go[w];
            if(p->val + 1 == q->val)
            {
                np->par = q;
            }
            else
            {
                State *nq = newState(p->val + 1);
                memcpy(nq->go, q->go, sizeof(q->go));
                nq->right = q->right;
                nq->par = q->par;
                q->par = nq;
                np->par = nq;
                while(p && p->go[w] == q)
                {
                    p->go[w] = nq;
                    p = p->par;
                }
            }
        }
        last = np;
    }
    void update()
    {
        State *tmp = last;
        while(tmp != root)
        {
            tmp->par->right++;
            tmp = tmp->par;
        }
    }
    int ask(char *s, int len)//询问这个串出现了几次
    {
        State *now = root;
        for(int i = 0; i < len; i++)
        {
            int w = s[i] - 'A';
            if(now->go[w])
                now = now->go[w];
            else
            {
                printf("0\n");
                return 0;
            }
        }
        printf("%d\n", now->right);
        return now->right;
    }
};

char s[maxl];

Suffix_Automation sam;

void decode(char *s, int len, int mask)
{
    for(int j = 0; j < len; j++)
    {
        mask = (mask*131 + j) % len;
        swap(s[j], s[mask]);
    }
    return;
}

int main()
{
    int Q;
    int mask = 0;
    scanf("%d", &Q);
    scanf("%s", s);
    int len = strlen(s);
    sam.init();
    for(int i = 0; i < len; i++)
        sam.extend(s[i] - 'A'), sam.update();
    while(Q--)
    {
        scanf("%s", s);
        switch(s[0])
        {
            case 'Q': scanf("%s", s);
                      len = strlen(s);
                      decode(s, len, mask);
                      mask ^= sam.ask(s, len);
                      break;
            case 'A': scanf("%s", s);
                      len = strlen(s);
                      decode(s, len, mask);
                      for(int i = 0; i < len; i++)
                          sam.extend(s[i] - 'A'), sam.update();
                      break;
        }
    }
    return 0;
}