【BZOJ3277】String (Generalized Suffix Automata)

Description

Given you n strings, ask how many substrings (excluding empty strings) of each string are substrings (including itself) of at least k of all n strings.

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Solution

Build a generalized suffix automaton to find out how many substrings each node is, denoted as size[u].
For each string, enumerate the right endpoints and find the satisfying size[u]>kpoint, len[u]that is, its contribution.

Time complexity is still a mystery. . .

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Code

/************************************************
 * Au: Hany01
 * Date: May 6th, 2018
 * Prob: [BZOJ3277] 串
 * Email: [email protected]
************************************************/

#include<bits/stdc++.h>

using namespace std;

typedef long long LL;
typedef pair<int, int> PII;
typedef set<int>::iterator It;
#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 x first
#define y second
#define pb(a) push_back(a)
#define mp(a, b) make_pair(a, b)
#define ALL(a) (a).begin(), (a).end()
#define SZ(a) ((int)(a).size())
#define INF (0x3f3f3f3f)
#define INF1 (2139062143)
#define Mod (1000000007)
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define y1 wozenmezhemecaia

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 int _, __; register char c_;
    for (_ = 0, __ = 1, c_ = getchar(); c_ < '0' || c_ > '9'; c_ = getchar()) if (c_ == '-') __ = -1;
    for ( ; c_ >= '0' && c_ <= '9'; c_ = getchar()) _ = (_ << 1) + (_ << 3) + (c_ ^ 48);
    return _ * __;
}

const int maxn = 100005;

int n, k, len[maxn << 1], sz[maxn << 1], ch[maxn << 1][26], fa[maxn << 1], v[maxn << 1], nex[maxn << 1], beg[maxn << 1], e, L[maxn], tot = 1, las;
vector<int> s[maxn];
set<int> Set[maxn << 1];
char str[maxn];

inline void extend(int c, int id)
{
    int np = ++ tot, p = las;
    las = tot, len[np] = len[p] + 1, Set[np].insert(id);
    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[np] = fa[q] = nq;
            while (p && ch[p][c] == q) ch[p][c] = nq, p = fa[p];
        }
    }
}

inline void add(int uu, int vv) { v[++ e] = vv, nex[e] = beg[uu], beg[uu] = e; }

void dfs(int u)
{
    for (register int i = beg[u]; i; i = nex[i]) {
        dfs(v[i]);
        while (Set[v[i]].begin() != Set[v[i]].end())
            Set[u].insert(*Set[v[i]].begin()), Set[v[i]].erase(Set[v[i]].begin());
    }
    sz[u] = SZ(Set[u]);
}

int main()
{
#ifdef hany01
    File("bzoj3277");
#endif

    n = read(), k = read();

    //Assert
    if (n < k) {
        rep(i, n) printf("0 ");
        return 0;
    }

    //Init
    For(i, 1, n) {
        scanf("%s", str), las = 1;
        rep(j, L[i] = strlen(str)) s[i].pb(str[j] - 97), extend(s[i][j], i);
    }

    //Build Tree
    For(i, 2, tot) add(fa[i], i);
    dfs(1);

    //Count
    For(i, 1, n) {
        LL Ans = 0; int u = 1;
        rep(j, L[i]) {
            u = ch[u][s[i][j]];
            while (sz[u] < k) u = fa[u];
            Ans += len[u];
        }
        printf("%lld ", Ans);
    }

    return 0;
}
//《马嵬坡》
//郑畋
//玄宗回马杨妃死，云雨难忘日月新。
//终是圣明天子事，景阳宫井又何人。