Codeforces 235 C

题目大意

给定一个模板串, 再给出(n)个询问, 询问每一个串的循环串总共在原串中出现了多少次.
循环串: 比如说有(str[] = {ABCD}), 则其循环串有({ABCD}, {BCDA}, {CDAB}, {DABC}), 共(len)个.

题解

把每一个串复制一遍放在原串后面: ({ABCD} o {ABCDABC}), 放入原串的后缀自动机中匹配. 在匹配时, 假如下一位无法匹配, 则跳suffix link; 假如即使跳了suffix link, 最大长度(len(suffix))仍然大于等于原串长度, 则也跳suffix link(相当于砍掉多余的部分).
放入SAM前要先跑一次KMP去循环节.

#include <cstdio>
#include <cstring>
#include <vector>
 
const int LEN = (int)1e6;
 
struct suffixAutomaton
{
    struct state
    {
        state *suc[26], *pre;
        int len;
        int sz, tg;
        std::vector<state*> bck;
 
        inline state()
        {
            for(int i = 0; i < 26; ++ i)
                suc[i] = NULL;
            pre = NULL;
            sz = 1;
            bck.clear();
            tg = 0;
        }
    };
 
    state *rt, *lst;
 
    inline void insert(int c)
    {
        state *u = new state;
        u->len = lst->len + 1;
        for(; lst != NULL && lst->suc[c] == NULL; lst->suc[c] = u, lst = lst->pre);
        if(lst == NULL)
            u->pre = rt;
        else
        {
            state *p = lst->suc[c];
            if(p->len == lst->len + 1)
                u->pre = p;
            else
            {
                state *q = new state;
                *q = *p;
                q->len = lst->len + 1, q->sz = 0;
                p->pre = u->pre = q;
                for(; lst != NULL && lst->suc[c] == p; lst->suc[c] = q, lst = lst->pre);
            }
        }
        lst = u;
    }
 
    void DFS(state *u)
    {
        u->tg = 1;
        if(u->pre != NULL)
            u->pre->bck.push_back(u);
        for(int i = 0; i < 26; ++ i)
            if(u->suc[i] != NULL && ! u->suc[i]->tg)
                DFS(u->suc[i]);
    }
 
    void get(state *u)
    {
        for(std::vector<state*>::iterator p = u->bck.begin(); p != u->bck.end(); ++ p)
            get(*p), u->sz += (*p)->sz;
    }
 
    inline void build(char *str, int len)
    {
        lst = rt = new state;
        rt->len = 0;
        for(int i = 0; i < len; ++ i)
            insert(str[i] - 'a');
        DFS(rt);
        get(rt);
    }
 
    inline int match(char *str, int len, int cir)
    {
        state *u = rt;
        int cur = 0;
        long long ans = 0;
        for(int i = 0; i < len + cir - 1; ++ i)
        {
            for(; u != rt && u->suc[str[i] - 'a'] == NULL; cur = u->pre->len, u = u->pre);
            if(u->suc[str[i] - 'a'] != NULL)
                u = u->suc[str[i] - 'a'], ++ cur;
            for(; u != rt && u->pre->len >= len; cur = u->pre->len, u = u->pre);
            if(cur >= len)
                ans += u->sz;
        }
        return ans;
    }
}SAM;
 
int main()
{
    #ifndef ONLINE_JUDGE
    freopen("CF235C.in", "r", stdin);
    #endif
    static char str[LEN];
    scanf("%s", str);
    int len = strlen(str);
    SAM.build(str, len);
    int n;
    scanf("%d
", &n);
    for(int i = 0; i < n; ++ i)
    {
        static char str[LEN << 1];
        scanf("%s", str);
        int len = strlen(str);
        static int nxt[LEN];
        nxt[0] = -1;
        int p = nxt[0];
        for(int i = 1; i < len; ++ i)
        {
            for(; ~ p && str[i] ^ str[p + 1]; p = nxt[p]);
            nxt[i] = str[i] == str[p + 1] ? ++ p : p;
        }
        int cir = len % (len - nxt[len - 1] - 1) == 0 ? len - nxt[len - 1] - 1 : len;
        for(int i = 0; i < cir; ++ i)
            str[i + len] = str[i];
        printf("%d
", SAM.match(str, len, cir));
    }
}