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
    
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      #include
      
       #include
       
        #include
        
         #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<
         
          <<' '; 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\n",l); }}int main(){ //freopen("in.in","r",stdin); n = read(); q = read(); scanf("%s",s + 1); getSA(); build(); solve(); return 0;}