题目
佳媛姐姐过生日的时候，她的小伙伴从某东上买了一个生日礼物。生日礼物放在一个神奇的箱子中。箱子外边写了
一个长为n的字符串s，和m个问题。佳媛姐姐必须正确回答这m个问题，才能打开箱子拿到礼物，升职加薪，出任CE
O，嫁给高富帅，走上人生巅峰。每个问题均有a,b,c,d四个参数，问你子串s[a..b]的所有子串和s[c..d]的最长公
共前缀的长度的最大值是多少？佳媛姐姐并不擅长做这样的问题，所以她向你求助，你该如何帮助她呢？
输入格式
输入的第一行有两个正整数n,m，分别表示字符串的长度和询问的个数。接下来一行是一个长为n的字符串。接下来
m行，每行有4个数a,b,c,d，表示询问s[a..b]的所有子串和s[c..d]的最长公共前缀的最大值。1<=n,m<=100,000,
字符串中仅有小写英文字母，a<=b,c<=d,1<=a,b,c,d<=n
输出格式
对于每一次询问，输出答案。
输入样例
5 5
aaaaa
1 1 1 5
1 5 1 1
2 3 2 3
2 4 2 3
2 3 2 4
输出样例
1
1
2
2
2
题解
一开始看错题，，原来是要求s[a...b]所有子串和子串s[c...d]的最大LCP【注意s[c...d]只是一个字符串】
那么我们二分一下答案
就需要快速判断子串c的height分组中是否存在子串s[a...b]
这个拿主席树记录位置存在信息就可以做到
height分组的边界可以倍增用ST表判断
还要注意不要越过子串的边界
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#define LL long long int
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define BUG(s,n) for (int i = 1; i <= (n); i++) cout<<s[i]<<' '; puts("");
using namespace std;
const int maxn = 200005,maxm = 6000005,INF = 1000000000;
inline int read(){
int out = 0,flag = 1; char c = getchar();
while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
return out * flag;
}
int n,q;
char s[maxn];
int sa[maxn],rank[maxn],height[maxn],t1[maxn],t2[maxn],bac[maxn],m;
void getSA(){
int *x = t1,*y = t2; m = 256;
for (int i = 0; i <= m; i++) bac[i] = 0;
for (int i = 1; i <= n; i++) bac[x[i] = s[i]]++;
for (int i = 1; i <= m; i++) bac[i] += bac[i - 1];
for (int i = n; i; i--) sa[bac[x[i]]--] = i;
for (int k = 1; k <= n; k <<= 1){
int p = 0;
for (int i = n - k + 1; i <= n; i++) y[++p] = i;
for (int i = 1; i <= n; i++) if (sa[i] - k > 0) y[++p] = sa[i] - k;
for (int i = 0; i <= m; i++) bac[i] = 0;
for (int i = 1; i <= n; i++) bac[x[y[i]]]++;
for (int i = 1; i <= m; i++) bac[i] += bac[i - 1];
for (int i = n; i; i--) sa[bac[x[y[i]]]--] = y[i];
swap(x,y);
p = x[sa[1]] = 1;
for (int i = 2; i <= n; i++)
x[sa[i]] = (y[sa[i]] == y[sa[i - 1]] && y[sa[i] + k] == y[sa[i - 1] + k] ? p : ++p);
if (p >= n) break;
m = p;
}
for (int i = 1; i <= n; i++) rank[sa[i]] = i;
for (int i = 1,k = 0; i <= n; i++){
if (k) k--;
int j = sa[rank[i] - 1];
while (s[i + k] == s[j + k]) k++;
height[rank[i]] = k;
}
}
int sum[maxm],ls[maxm],rs[maxm],rt[maxn],cnt;
void add(int& u,int pre,int l,int r,int pos){
sum[u = ++cnt] = sum[pre] + 1;
ls[u] = ls[pre]; rs[u] = rs[pre];
if (l == r) return;
int mid = l + r >> 1;
if (mid >= pos) add(ls[u],ls[pre],l,mid,pos);
else add(rs[u],rs[pre],mid + 1,r,pos);
}
int query(int u,int v,int l,int r,int L,int R){
if (l >= L && r <= R) return sum[u] - sum[v];
int mid = l + r >> 1;
if (mid >= R) return query(ls[u],ls[v],l,mid,L,R);
if (mid < L) return query(rs[u],rs[v],mid + 1,r,L,R);
return query(ls[u],ls[v],l,mid,L,R) + query(rs[u],rs[v],mid + 1,r,L,R);
}
int a,b,c,d;
int bin[maxn],Log[maxn],mn[maxn][20];
void build(){
for (int i = 1; i <= n; i++) add(rt[i],rt[i - 1],1,n,sa[i]);
bin[0] = 1;
for (int i = 1; i <= 20; i++) bin[i] = bin[i - 1] << 1;
Log[0] = -1;
for (int i = 1; i < maxn; i++) Log[i] = Log[i >> 1] + 1;
for (int i = 1; i <= n; i++) mn[i][0] = height[i];
for (int i = 1; i <= 17; i++)
for (int j = 1; j <= n; j++){
if (j + bin[i] - 1 > n) break;
mn[j][i] = min(mn[j][i - 1],mn[j + bin[i - 1]][i - 1]);
}
}
int getm(int l,int r){
if (l > r) return INF;
int t = Log[r - l + 1];
return min(mn[l][t],mn[r - bin[t] + 1][t]);
}
bool check(int len){
if (len == 0) return true;
int u = rank[c],tmp,l = u,r = u;
for (int i = 17; i >= 0; i--){
tmp = l - bin[i];
if (tmp > 0 && getm(tmp + 1,u) >= len) l = tmp;
tmp = r + bin[i];
if (tmp <= n && getm(u + 1,tmp) >= len) r = tmp;
}
return query(rt[r],rt[l - 1],1,n,a,b - len + 1);
}
void solve(){
while (q--){
a = read(); b = read();
c = read(); d = read();
int l = 0,r = min(d - c + 1,b - a + 1),mid;
while (l < r){
mid = l + r + 1 >> 1;
if (check(mid)) l = mid;
else r = mid - 1;
}
printf("%d
",l);
}
}
int main(){
//freopen("in.in","r",stdin);
n = read(); q = read();
scanf("%s",s + 1);
getSA();
build();
solve();
return 0;
}