[HEOI/TJOI2016][洛谷P4094]字符串(SA+主席树)
题面
https://www.luogu.com.cn/problem/P4094
题解
前置知识：
题目给出字符串S，每次询问给出a,b,c,d，问S[a..b]的所有子串与S[c..d]的lcp的最大值。
考虑简化题意：S[a..b]的所有子串->S[a..b]的所有后缀，因为子串的右端点往右移肯定不会变差。
首先预处理出后缀数组和height，以及height的ST表。再以0为下界，\(\min(d-c+1, b-a+1)\)为上界二分答案。假设当前答案A成立，需满足\(\exist x \in[a,b-A+1]{\ } s.t.{\ }lcp(suf_x,suf_c) {\geq} A\)。而满足\(lcp(suf_x,suf_c) \geq A\)的x的rank值是连续的一个区间（rank是后缀数组里的那个），可以二分得出这个区间的左右端点l、r。
接下来的任务是：判断\(rank[a],rank[a+1],…,rank[b-A+1]\)这些值里面，有没有\(\in [l,r]\)的。这个就可以使用主席树解决。
总时间复杂度\(O(n \log^2n)\)。
代码
#include<bits/stdc++.h>
using namespace std;
#define rg register
#define In inline
const int N = 1e5;
const int TN = 2e6;
In int read(){
int s = 0,ww = 1;
char ch = getchar();
while(ch < '0' || ch > '9'){if(ch == '-')ww = -1;ch = getchar();}
while('0' <= ch && ch <= '9'){s = 10 * s + ch - '0';ch = getchar();}
return s * ww;
}
In void write(int x){
if(x < 0)putchar('-'),x = -x;
if(x > 9)write(x / 10);
putchar('0' + x % 10);
}
int n,m;
char s[N+5];
int lg[N+5];
struct CMTree{
int rt[N+5],lc[TN+5],rc[TN+5],num[TN+5];
int cnt,rn;
In void pushup(int u){
num[u] = num[lc[u]] + num[rc[u]];
}
void ud(int u1,int u2,int l,int r,int d){
if(l == r){
num[u2]++;
return;
}
int m = (l + r) >> 1;
if(d <= m){
int v = ++cnt;
lc[u2] = v,rc[u2] = rc[u1];
ud(lc[u1],lc[u2],l,m,d);
}
else{
int v = ++cnt;
rc[u2] = v,lc[u2] = lc[u1];
ud(rc[u1],rc[u2],m + 1,r,d);
}
pushup(u2);
}
int query(int u1,int u2,int l,int r,int ql,int qr){
if(l == ql && r == qr)return num[u2] - num[u1];
int m = (l + r) >> 1;
if(qr <= m)return query(lc[u1],lc[u2],l,m,ql,qr);
else if(ql > m)return query(rc[u1],rc[u2],m + 1,r,ql,qr);
else return query(lc[u1],lc[u2],l,m,ql,m) + query(rc[u1],rc[u2],m + 1,r,m + 1,qr);
}
void insert(int x){
rt[++rn] = ++cnt;
ud(rt[rn-1],rt[rn],0,n,x);
}
int sum(int a,int b,int ql,int qr){
return query(rt[a-1],rt[b],0,n,ql,qr);
}
}T;
struct ST{
int minn[N+5][18];
void prepro(int a[]){
for(rg int i = 1;i <= n;i++)minn[i][0] = a[i];
for(rg int j = 1;j <= 17;j++){
for(rg int i = 1;i + (1<<j) - 1 <= n;i++)minn[i][j] = min(minn[i][j-1],minn[i+(1<<(j-1))][j-1]);
}
}
int query(int l,int r){
int d = lg[r-l+1];
return min(minn[l][d],minn[r+1-(1<<d)][d]);
}
};
struct SA{
int sa[N+5],temp[N+5],rk[N+5],h[N+5],num[N+5];
int m;
ST H;
void qsort(){
memset(num,0,sizeof(int) * (m+1));
for(rg int i = 1;i <= n;i++)num[rk[i]]++;
for(rg int i = 2;i <= m;i++)num[i] += num[i-1];
for(rg int i = n;i >= 1;i--)sa[num[rk[temp[i]]]--] = temp[i];
}
void calch(){
int k = 0;
for(rg int i = 1;i <= n;i++){
if(rk[i] == 1)h[1] = k = 0;
else{
if(k)k--;
int j = sa[rk[i]-1];
while(s[i+k] == s[j+k])k++;
h[rk[i]] = k;
}
}
}
void init(){
m = 26;
for(rg int i = 1;i <= n;i++)temp[i] = i,rk[i] = s[i] - 'a' + 1;
qsort();
for(rg int d = 1;d <= n;d <<= 1){
int cnt = 0;
for(rg int i = n - d + 1;i <= n;i++)temp[++cnt] = i;
for(rg int i = 1;i <= n;i++)if(sa[i] > d)temp[++cnt] = sa[i] - d;
qsort();
memcpy(temp,rk,sizeof(int) * (n+1));
cnt = rk[sa[1]] = 1;
for(rg int i = 2;i <= n;i++){
if(temp[sa[i]] != temp[sa[i-1]] || temp[sa[i]+d] != temp[sa[i-1]+d])cnt++;
rk[sa[i]] = cnt;
}
if(cnt == n)break;
m = cnt;
}
calch();
H.prepro(h);
for(rg int i = 1;i <= n;i++)T.insert(rk[i]);
}
int lcp(int i,int j){
if(i == j)return n - i + 1;
int x = rk[i],y = rk[j];
if(x > y)swap(x,y);
return H.query(x + 1,y);
}
}S;
In bool check(int A,int a,int b,int c){
if(!A)return 1;
int L,R;
L = 1,R = S.rk[c];
while(L < R){
int mid = (L + R) >> 1;
if(S.lcp(S.sa[mid],c) >= A)R = mid;
else L = mid + 1;
}
int l = L;
L = S.rk[c],R = n;
while(L < R){
int mid = (L + R + 1) >> 1;
if(S.lcp(c,S.sa[mid]) >= A)L = mid;
else R = mid - 1;
}
int r = R;
return T.sum(a,b - A + 1,l,r);
}
int main(){
for(rg int i = 2;i <= N;i++)lg[i] = lg[i>>1] + 1;
n = read(); m = read();
scanf("%s",s + 1);
S.init();
while(m--){
int a = read(),b = read(),c = read(),d = read();
int L = 0,R = min(b - a + 1,d - c + 1);
while(L < R){
int mid = (L + R + 1) >> 1;
if(check(mid,a,b,c))L = mid;
else R = mid - 1;
}
write(L),putchar('\n');
}
return 0;
}
[]
