【BZOJ2780】【SPOJ】Sevenk Love Oimaster（广义后缀自动机）

Description

有n个大串和m个询问，每次给出一个字符串s询问在多少个大串中出现过。

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Solution

广义后缀自动机入门题。

其实就是在插入一个串后将last设为root即可。
然后统计每个节点所代表的串是多少个输入串的子串。
对于每次询问，直接在SAM上走一遍就行了。

至于时间复杂度，感觉真的很迷啊。。

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Code

/**************************************
 * Au: Hany01
 * Date: May 5th, 2018
 * Prob: [BZOJ2780][SPOJ JZPGYZ] Sevenk Love Oimaster
 * Email: [email protected]
**************************************/

#include<bits/stdc++.h>

using namespace std;

typedef long long LL;
typedef pair<int, int> PII;
typedef vector<int> VI;
#define File(a) freopen(a".in", "r", stdin), freopen(a".out", "w", stdout)
#define rep(i, j) for (register int i = 0, i##_end_ = j; i < i##_end_; ++ i)
#define For(i, j ,k) for (register int i = (j), i##_end_ = (k); i <= i##_end_; ++ i)
#define Fordown(i, j, k) for (register int i = (j), i##_end_ = (k); i >= i##_end_; -- i)
#define Set(a, b) memset(a, b, sizeof(a))
#define Cpy(a, b) memcpy(a, b, sizeof(a))
#define SZ(a) ((int)(a.size()))
#define ALL(a) a.begin(), a.end()
#define pb(a) push_back(a)
#define mp(a, b) make_pair(a, b)
#define INF (0x3f3f3f3f)
#define INF1 (2139062143)
#define Mod (1000000007)
#define y1 wozenmezhemecaia 
#ifdef hany01
#define debug(...) fprintf(stderr, __VA_ARGS__)
#else
#define debug(...)
#endif

template<typename T> inline bool chkmax(T &a, T b) { return a < b ? a = b, 1 : 0; }
template<typename T> inline bool chkmin(T &a, T b) { return b < a ? a = b, 1 : 0; }

inline int read() {
    register char c_; register int _, __;
    for (_ = 0, __ = 1, c_ = getchar(); !isdigit(c_); c_ = getchar()) if (c_ == '-')  __ = -1;
    for ( ; isdigit(c_); c_ = getchar()) _ = (_ << 1) + (_ << 3) + (c_ ^ 48);
    return _ * __;
}

const int maxn = 10005, maxl = 100005, maxq = 360005;

int tot = 1, las, len[maxl << 1], fa[maxl << 1], ch[maxl << 1][26], mk[maxl << 1], cnt[maxl << 1], n, m, L[maxn];
char p[maxq];
string s[maxn];

inline void extend(int c)
{
    int np = ++ tot, p = las;
    las = tot, len[np] = len[p] + 1;
    while (p && !ch[p][c]) ch[p][c] = np, p = fa[p];
    if (!p) fa[np] = 1;
    else {
        int q = ch[p][c];
        if (len[q] == len[p] + 1) fa[np] = q;
        else {
            int nq = ++ tot;
            fa[nq] = fa[q], Cpy(ch[nq], ch[q]), len[nq] = len[p] + 1;
            fa[q] = fa[np] = nq;
            while (p && ch[p][c] == q) ch[p][c] = nq, p = fa[p];
        }
    }
}

int main()
{
#ifdef hany01
    File("bzoj2780");
#endif

    ios::sync_with_stdio(false);

    cin >> n >> m;
    For(i, 1, n) {
        cin >> s[i];
        L[i] = s[i].length(), las = 1;
        rep(j, L[i]) extend(s[i][j] - 97);
    }

    For(i, 1, n) {
        int u = 1;
        rep(j, L[i]) {
            u = ch[u][s[i][j] - 97];
            int v = u;
            while (v && mk[v] != i) mk[v] = i, ++ cnt[v], v = fa[v];
        }
    }

    while (m --)
    {
        cin >> p;
        int u = 1;
        rep(i, strlen(p)) u = ch[u][p[i] - 97];
        printf("%d\n", cnt[u]);
    }

    return 0;
}
//露清枕簟藕花香，恨悠扬。
//    -- 顾敻《虞美人·触帘风送景阳钟》