BZOJ 4698: Sdoi2008 Sandy的卡片

后缀数组模板题，很久之前写过。

$sa[i]$ 表示排名为 $i$ 的后缀的位置
$rank[i]$ 表示后缀 $i$ 的排名
$height[i]$ 表示 $sa[i]$ 与 $sa[i-1]$ 的最长公共前缀，即排名相邻的两个后缀的LCP

#include <cstdio>

const int N = 102005, INF = 0x3f3f3f3f;

int n, m, t;
int a[1005][1005], sa[N], c[N], x[N], y[N], s[N];
int rank[N], height[N], vis[N], sta[N], top, id[N];

int read() {
    int x = 0, f = 1;
    char c = getchar();
    while (c < '0' || c > '9') {
        if (c == '-') f = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9') {
        x = (x << 3) + (x << 1) + (c ^ 48);
        c = getchar();
    }
    return x * f;
}
int min(int x, int y) {
    if (x <= y) return x; return y;
}
int max(int x, int y) {
    if (x >= y) return x; return y;
}
void swap(int &x, int &y) {
    x ^= y, y ^= x, x ^= y;
}

void get_sa() {
    for (int i = 0; i < n; ++i) ++c[x[i] = s[i]];
    for (int i = 1; i < m; ++i) c[i] += c[i-1];
    for (int i = n - 1; i >= 0; --i) sa[--c[x[i]]] = i;

    for (int k = 1; k <= n; k <<= 1) {
        int p = 0;
        for (int i = n - k; i < n; ++i) y[p++] = i;
        for (int i = 0; i < n; ++i) if (sa[i] >= k) y[p++] = sa[i] - k;

        for (int i = 0; i < m; ++i) c[i] = 0;
        for (int i = 0; i < n; ++i) ++c[x[y[i]]];
        for (int i = 1; i < m; ++i) c[i] += c[i-1];
        for (int i = n - 1; i >= 0; --i) sa[--c[x[y[i]]]] = y[i];

        for (int i = 0; i < n; ++i) swap(x[i], y[i]);

        p = 1, x[sa[0]] = 0;
        for (int i = 1; i < n; ++i)
            x[sa[i]] = y[sa[i]] == y[sa[i-1]] && (sa[i]+k<n ? y[sa[i]+k] : -1) == (sa[i-1]+k<n ? y[sa[i-1]+k] : -1) ? p-1 : p++;
        if (p >= n) break; m = p;
    }
} 
void get_height() {
    int k = 0;
    for (int i = 0; i < n; ++i) {
        if (!x[i]) continue;
        if (k) --k;
        int j = sa[x[i]-1];
        while (j+k<n && i+k<n && s[j+k]==s[i+k]) ++k;
        height[x[i]] = k;
    }
}
bool check(int x) {
    while (top) vis[sta[top--]] = 0;
    for (int i = 0; i < n; ++i) {
        if (height[i] < x) while(top) vis[sta[top--]] = 0;
        if (!vis[id[sa[i]]]) {
            vis[id[sa[i]]] = 1;
            sta[++top] = id[sa[i]];
            if (top == t) return true;
        }
    }
    return false;
}

int main() {
    t = read(); int l = 0, r = INF, minn = INF, maxn = 0;
    for (int i = 1; i <= t; ++i) {
        a[i][0] = read(), a[i][1] = read();
        r = min(r, a[i][0] - 1);
        for (int j = 2; j <= a[i][0]; ++j) {
            a[i][j] = read();
            minn = min(minn, a[i][j] - a[i][j-1]);
            maxn = max(maxn, a[i][j] - a[i][j-1]);
        }
    }
    for (int i = 1; i <= t; ++i) {
        for (int j = 2; j <= a[i][0]; ++j)
            id[n] = i, s[n++] = a[i][j] - a[i][j-1] - minn;
        s[n++] = ++maxn-minn;
    }
    for (int i = 0; i < n; ++i) m = max(m, s[i] + 1);
    get_sa(), get_height();
    while (l < r) {
        int mid = l + ((r - l + 1) >> 1);
        if (check(mid)) l = mid; else r = mid - 1;
    }
    printf("%d\n", l + 1);
    return 0;
}