Description

click me

Solution

一道不那么裸的SAM

显然先要二分L，接下来的问题是怎么判断是否可行。
先用sam求出从每个位置最多可以匹配多长。考虑一个DP，dp[i]表示到第i个位置最多可以匹配多少位置，转移非常显然，用单调队列优化就行了。

Code

/************************************************
 * Au: Hany01
 * Date: May 7th, 2018
 * Prob: [BZOJ2806][CTSC2012] 熟悉的文章
 * Email: [email protected]
************************************************/

#include<bits/stdc++.h>

using namespace std;

typedef long long LL;
typedef pair<int, int> PII;
#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 = 1100005;

int tot = 1, las, len[maxn << 1], fa[maxn << 1], ch[maxn << 1][2], lens, mxl[maxn], dp[maxn], q[maxn], head, tail, n, m;
char 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;
            Cpy(ch[nq], ch[q]), fa[nq] = fa[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 init()
{
    int u = 1, mx = 0;
    rep(i, lens) {
        int c = s[i] ^ 48;
        if (ch[u][c]) u = ch[u][c], ++ mx;
        else {
            while (u && !ch[u][c]) u = fa[u];
            if (!u) u = 1, mx = 0;
            else mx = len[u] + 1, u = ch[u][c];
        }
        mxl[i + 1] = mx;
    }
}

inline int DP(int L)
{
    head = 1, tail = 0;
    For(i, 1, lens) {
        dp[i] = dp[i - 1];
        if (i < L) continue;
        while (head <= tail && dp[q[tail]] + i - q[tail] <= dp[i - L] + i - (i - L)) -- tail;
        q[++ tail] = i - L;
        while (head <= tail && q[head] < i - mxl[i]) ++ head;
        if (head <= tail) chkmax(dp[i], dp[q[head]] + i - q[head]);
    }
    return dp[lens];
}

inline void BS()
{
    int l = 0, r = lens, mid;
    while (l < r) {
        mid = (l + r + 1) >> 1;
        if (DP(mid) * 10 >= lens * 9) l = mid; else r = mid - 1;
    }
    printf("%d\n", l);
}

int main()
{
#ifdef hany01
    File("bzoj2806");
#endif

    for (n = read(), m = read(); m --; ) {
        scanf("%s", s), las = 1;
        rep(j, strlen(s)) extend(s[j] ^ 48);
    }

    while (n --)
        scanf("%s", s), lens = strlen(s), init(), BS();

    return 0;
}
//玉惨花愁出凤城，莲花楼下柳青青。
//    -- 聂胜琼《鹧鸪天·别情》