P4094 [HEOI2016/TJOI2016]字符串

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求$s[l_1..r_1]$中所有子串和$s[l_2..r_2]$的最长公共前缀。

模版题。看清楚题目，答案具有单调性，二分长度，利用倍增和线段树合并查询下是否存在子串即可。

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 = 2e5 + 10;
const int mod = 1e9 + 7;

const int SN = 2e5 + 100;
int n;
struct seg
{
    int l, r, sum;
} tree[N * 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];
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 p)
    {
        int cur = ++scnt;
        mxl[cur] = mxl[pre] + 1;
        int u;
        modify(rt[cur], 1, n, p);
        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], f[SN][21];
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 = 1; i <= 20; i++)
        f[x][i] = f[f[x][i - 1]][i - 1];
    for (int i = head[x]; i; i = e[i].nxt)
    {

        int to = e[i].to;
        f[to][0] = x;
        dfs(to);
        rt[x] = merge(rt[x], rt[to], 1, n);
    }
}

char s[N];
int ed[N];
int m;

bool check(int l, int r, int ll, int rr, int x)
{
    if (x == 0)
        return 1;
    int u = ed[ll + x - 1];
    //cout << u << endl;
    for (int i = 20; i >= 0; i--)
        if (f[u][i] && sam.mxl[f[u][i]] >= x)
            u = f[u][i];
    //cout << u << endl;
    return query(rt[u], l + x - 1, r, 1, n) > 0;
}
int main()
{
    scanf("%d%d", &n, &m);
    scanf("%s", s + 1);
    for (int i = 1; i <= n; i++)
        sam.insert(s[i] - 'a', i), ed[i] = sam.pre;
    for (int i = 2; i <= sam.scnt; i++)
        addadge(sam.link[i], i);
    ///out << sam.scnt << endl;

    dfs(1);

    while (m--)
    {
        int l1, r1, l2, r2;
        scanf("%d%d%d%d", &l1, &r1, &l2, &r2);
        int l = 0;
        int r = min(r2 - l2 + 1, r1 - l1 + 1);
        int ans = 0;
        // cout << "---" << endl;
        // cout << check(l1, r1, l2, r2, 0) << endl;
        while (l <= r)
        {
            //cout << l << " " << r << endl;
            int mid = (l + r) >> 1;
            if (check(l1, r1, l2, r2, mid))
            {
                ans = mid;
                l = mid + 1;
            }
            else
                r = mid - 1;
        }
        printf("%d\n", ans);
    }
}