嘟嘟嘟
今天复习一下SAM。
lcp固然不好做，干脆直接翻过来变成后缀。首先答案一定满足单调性，所以我们二分lcp的长度(mid)，然后判断(s[d ldots d + mid - 1])是否在(s[b ldots a])（别忘了整个串是反过来的）中出现即可。
怎么判断是否出现呢？其实就是判断这个子串的endpos是否在(s[b + mid - 1 ldots a])中，因此我们要求出SAM上的每一个节点的endpos集合，这就要用到线段树合并了。
需要注意的是，并不是直接在(d)在SAM上的节点的线段树开始找，需要一直往上跳祖先，直到满足这个节点的len最小，且仍(geqslant mid)。因为这样endpos集合的元素就会更多，找到的概率就更大。
线段树合并没有垃圾回收，不过出题人比较良心，不卡。
#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
#define enter puts("")
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define In inline
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 1e5 + 5;
const int N = 20;
inline ll read()
{
ll ans = 0;
char ch = getchar(), last = ' ';
while(!isdigit(ch)) last = ch, ch = getchar();
while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
if(last == '-') ans = -ans;
return ans;
}
inline void write(ll x)
{
if(x < 0) x = -x, putchar('-');
if(x >= 10) write(x / 10);
putchar(x % 10 + '0');
}
char s[maxn];
int n, m, len;
struct Tree
{
int ls, rs, sum;
}t[maxn * N * N];
int root[maxn << 1], cur[maxn << 1], tcnt = 0;
In void update(int& now, int l, int r, int id)
{
if(!now) now = ++tcnt;
if(l == r) {++t[now].sum; return;}
int mid = (l + r) >> 1;
if(id <= mid) update(t[now].ls, l, mid, id);
else update(t[now].rs, mid + 1, r, id);
t[now].sum = t[t[now].ls].sum + t[t[now].rs].sum;
}
In int merge(int x, int y, int l, int r)
{
if(!x || !y) return x | y;
if(l == r) {t[x].sum += t[y].sum; return x;}
int mid = (l + r) >> 1, z = ++tcnt;
t[z].ls = merge(t[x].ls, t[y].ls, l, mid);
t[z].rs = merge(t[x].rs, t[y].rs, mid + 1, r);
t[z].sum = t[t[z].ls].sum + t[t[z].rs].sum;
return z;
}
In int query(int now, int L, int R, int l, int r)
{
if(!now) return 0;
if(L == l && R == r) return t[now].sum;
int mid = (l + r) >> 1;
if(R <= mid) return query(t[now].ls, L, R, l, mid);
else if(L > mid) return query(t[now].rs, L, R, mid + 1, r);
else return query(t[now].ls, L, mid, l, mid) + query(t[now].rs, mid + 1, R, mid + 1, r);
}
struct Sam
{
int las, cnt;
int tra[maxn << 1][27], len[maxn << 1], link[maxn << 1];
In void init() {link[las = cnt = 0] = -1;}
In void insert(int c)
{
int now = ++cnt, p = las;
len[now] = len[las] + 1;
while(~p && !tra[p][c]) tra[p][c] = now, p = link[p];
if(p == -1) link[now] = 0;
else
{
int q = tra[p][c];
if(len[q] == len[p] + 1) link[now] = q;
else
{
int clo = ++cnt;
memcpy(tra[clo], tra[q], sizeof(tra[q]));
len[clo] = len[p] + 1;
link[clo] = link[q], link[q] = link[now] = clo;
while(~p && tra[p][c] == q) tra[p][c] = clo, p = link[p];
}
}
las = now;
}
int buc[maxn << 1], pos[maxn << 1];
In void solve()
{
for(int i = 1; i <= cnt; ++i) ++buc[len[i]];
for(int i = 1; i <= cnt; ++i) buc[i] += buc[i - 1];
for(int i = 1; i <= cnt; ++i) pos[buc[len[i]]--] = i;
for(int i = cnt; i; --i)
{
int now = pos[i], fa = link[now];
root[fa] = merge(root[fa], root[now], 0, n - 1);
}
}
}S;
int fa[N + 2][maxn << 1];
In bool judge(int len, int x, int L, int R)
{
for(int i = 20; i >= 0; --i)
if(fa[i][x] && S.len[fa[i][x]] >= len) x = fa[i][x];
return query(root[x], L + len - 1, R, 0, n - 1);
}
int main()
{
//freopen("ha.in", "r", stdin);
//freopen("ha.out", "w", stdout);
n = read(), m = read();
scanf("%s", s);
len = strlen(s); S.init();
reverse(s, s + len);
for(int i = 0; i < n; ++i)
{
S.insert(s[i] - 'a'); cur[i] = S.las;
update(root[cur[i]], 0, n - 1, i);
}
S.solve();
for(int i = 1; i <= S.cnt; ++i) fa[0][i] = S.link[i];
for(int j = 1; j <= N; ++j)
for(int i = 1; i <= S.cnt; ++i) fa[j][i] = fa[j - 1][fa[j - 1][i]];
for(int i = 1; i <= m; ++i)
{
int a = n - read(), b = n - read(), c = n - read(), d = n - read();
int L = 0, R = min(a - b + 1, c - d + 1);
while(L < R)
{
int mid = (L + R + 1) >> 1;
if(judge(mid, cur[c], b, a)) L = mid;
else R = mid - 1;
}
write(L), enter;
}
return 0;
}