洛谷3804 【模板】后缀自动机

题目

给定一个只包含小写字母的字符串SS,

请你求出 SS 的所有出现次数不为 11 的子串的出现次数乘上该子串长度的最大值。

输入格式

一行一个仅包含小写字母的字符串SS

输出格式

一个整数，为 所求答案

输入样例

abab

输出样例

4

提示

对于\(10\%10%\)的数据，|S|<=1000∣S∣<=1000
对于\(100\%100%\)的数据，|S|<=\(10^6\)∣S∣<=\(10^6\)

题解

贴个模板。。。
这篇讲得不错
lible的讲解

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define LL long long int
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define Redge(u) for (int k = h[u]; k; k = ed[k].nxt)
using namespace std;
const int maxn = 2000005,maxm = 100005,INF = 1000000000;
inline int RD(){
	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 << 1) + (out << 3) + c - '0'; c = getchar();}
	return out * flag;
}
int ch[maxn][26],pre[maxn],step[maxn],n,cnt,last;
int b[maxn],sz[maxn],a[maxn];
LL ans = 0;
char s[maxn];
void ins(int u){
	int p = last,np = ++cnt;
	last = np; step[np] = step[p] + 1;
	while (p && !ch[p][u]) ch[p][u] = np,p = pre[p];
	if (!p) pre[np] = 1;
	else {
		int q = ch[p][u];
		if (step[q] == step[p] + 1) pre[np] = q;
		else {
			int nq = ++cnt; step[nq] = step[p] + 1;
			for (int i = 0; i < 26; i++) ch[nq][i] = ch[q][i];
			pre[nq] = pre[q]; pre[q] = pre[np] = nq;
			while (ch[p][u] == q) ch[p][u] = nq,p = pre[p];
		}
	}
	sz[np] = 1;
}
void solve(){
	REP(i,cnt) b[step[i]]++;
	REP(i,cnt) b[i] += b[i - 1];
	REP(i,cnt) a[b[step[i]]--] = i;
	for (int i = cnt; i; i--){
		sz[pre[a[i]]] += sz[a[i]];
		if (sz[a[i]] > 1) ans = max(ans,1ll * step[a[i]] * sz[a[i]]);
	}
}
int main(){
	scanf("%s",s + 1);
	cnt = last = 1; n = strlen(s + 1);
	for (int i = 1; i <= n; i++) ins(s[i] - 'a');
	solve();
	printf("%lld",ans);
	return 0;
}