CF1037H. Security

发表于 | 阅读数

给出一个字符串$S$,给出$Q$个操作，给出$L, R, T$，求字典序最小的$S[l,r]$的子串,并且它的字典序严格大于$T$如果无解输出$−1$

贪心的去考虑一定是$T$的前缀（包括空串）+一个字母

建立$SAM$，暴力地跑前缀，如果前缀没有出现在$[l,r]$区间内，则$break$。否则对所有存在的前缀暴力枚举字母$[t[i+1]-‘a’+1,’z]$，而且越是长的前缀+字母越优。

  • 考虑$endpos$是表示集合里$S$出现的位置。即$trans[u][id]$搜到$endpos$的时候此时$S$出现的位置为$endpos$。初始化每个前缀出现的位置加到对应的点上，后缀链接建立树，树上启发式合并。预处理即可。
代码 
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#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define pii pair<int, int>
#define mk make_pair
#define ls tree[pos].l
#define rs tree[pos].r
const int N = 1e6 + 10;
const int mod = 1e9 + 7;

const int SN = 2e5 + 100;

struct seg
{
    int l, r, sum;
} tree[SN * 20];
int scnt;
void pushup(int pos)
{
    tree[pos].sum = tree[ls].sum + tree[rs].sum;
}
void modify(int &pos, int l, int r, int w)
{
    if (!pos)
        pos = ++scnt;
    if (l == r)
    {
        tree[pos].sum++;
        return;
    }

    int mid = (l + r) >> 1;
    if (w <= mid)
        modify(ls, l, mid, w);
    else
        modify(rs, mid + 1, r, w);
    pushup(pos);
}
int query(int pos, int ql, int qr, int l, int r)
{
    if (ql <= l && r <= qr)
        return tree[pos].sum;
    int mid = (l + r) >> 1;
    int res = 0;
    if (ql <= mid)
        res += query(ls, ql, qr, l, mid);
    if (qr > mid)
        res += query(rs, ql, qr, mid + 1, r);
    return res;
}

int merge(int u, int v, int l, int r)
{
    if (!u || !v)
        return u + v;
    int pos = ++scnt;
    if (l == r)
    {
        tree[pos].sum = tree[u].sum + tree[v].sum;
        return pos;
    }
    int mid = (l + r) >> 1;
    ls = merge(tree[u].l, tree[v].l, l, mid);
    rs = merge(tree[u].r, tree[v].r, mid + 1, r);
    pushup(pos);
    return pos;
}

int rt[SN], n;
struct SAM
{
    int trans[SN][26];
    int mxl[SN], link[SN], pre, scnt;
    SAM() { pre = scnt = 1; };
    void init()
    {
        for (int j = 1; j <= scnt; j++)
        {
            link[j] = 0;
            mxl[j] = 0;
            //siz[j] = 0;
            memset(trans[j], 0, sizeof(trans[j]));
        }
        scnt = pre = 1;
    };
    void insert(int id)
    {
        int cur = ++scnt;
        mxl[cur] = mxl[pre] + 1;
        int u;
        modify(rt[cur], 1, n, mxl[cur]);
        for (u = pre; u && !trans[u][id]; u = link[u])
            trans[u][id] = cur;
        pre = cur;
        if (!u)
            link[cur] = 1;
        else
        {
            int x = trans[u][id];
            if (mxl[x] == mxl[u] + 1)
                link[cur] = x;
            else
            {
                int nc = ++scnt;
                link[nc] = link[x];
                mxl[nc] = mxl[u] + 1;
                memcpy(trans[nc], trans[x], sizeof(trans[x]));

                link[cur] = link[x] = nc;
                for (; u && trans[u][id] == x; u = link[u])
                    trans[u][id] = nc;
            }
        }
    }
} sam;

struct edge
{
    int to, nxt;
} e[SN];
int ecnt, head[SN];
void addadge(int u, int v)
{
    e[++ecnt].to = v;
    e[ecnt].nxt = head[u];
    head[u] = ecnt;
}
void dfs(int x)
{
    for (int i = head[x]; i; i = e[i].nxt)
    {
        int to = e[i].to;
        dfs(to);
        rt[x] = merge(rt[x], rt[to], 1, n);
    }
}
void solve(int l, int r, char *t)
{
    int u = 1;
    int len = strlen(t + 1);
    t[len + 1] = 'a' - 1;
    int ansi = -1, ansj = -1;
    for (int i = 1; i <= len + 1; i++)
    {
        //cout << t[i] << "---" << endl;
        for (int j = t[i] - 'a' + 1; j < 26; j++)
        {
            int v = sam.trans[u][j];
            if (v && query(rt[v], l + i - 1, r, 1, n)) // ?
            {
                //cout << i << " " << j << endl;
                ansi = i;
                ansj = j;
                break;
            }
        }
        if (i == len + 1)
            break;
        u = sam.trans[u][t[i] - 'a'];
        if (!u || !query(rt[u], l + i - 1, r, 1, n))
            break;
    }
    if (ansi == -1)
        printf("-1\n");
    else
    {
        // cout << ansi << endl;
        for (int i = 1; i <= ansi - 1; i++)
            printf("%c", t[i]);
        printf("%c", ansj + 'a');
        printf("\n");
    }
}

char s[N], t[N];
int q;
int main()
{
    scanf("%s", s + 1);
    n = strlen(s + 1);
    for (int i = 1; i <= n; i++)
        sam.insert(s[i] - 'a');
    for (int i = 2; i <= sam.scnt; i++)
        addadge(sam.link[i], i);
    //cout << sam.scnt << endl;
    dfs(1);
    scanf("%d", &q);
    while (q--)
    {
        int l, r;
        scanf("%d%d%s", &l, &r, t + 1);
        solve(l, r, t);
    }
}