luogu3804 【模板】后缀自动机

题目链接

solution

用后缀数组水一发后缀自动机的模板233

先跑一遍后缀数组，求出来height。然后根据height的性质，枚举造成贡献的位置，然后看一下这个位置在最长多长的一段区间内均为最小值。答案就是(len_i*height_i)。(len_i)表示最长长度为(len_i)的一段区间(height)的最小值都是(height_i)。用单调栈统计即可。

code

/*
* @Author: wxyww
* @Date:   2020-04-20 07:58:21
* @Last Modified time: 2020-04-20 08:36:41
*/
#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<queue>
#include<vector>
#include<ctime>
using namespace std;
typedef long long ll;
const int N = 1000010;
ll read() {
	ll x = 0,f = 1;char c = getchar();
	while(c < '0' || c > '9') {
		if(c == '-') f = -1; c = getchar();
	}
	while(c >= '0' && c <= '9') {
		x = x * 10 + c - '0'; c = getchar();
	}
	return x * f;
}
char s[N];
int sa[N],height[N],c[N],x[N],y[N],n,m;
void get_sa() {
	for(int i = 1;i <= n;++i) c[x[i] = s[i]]++;
	for(int i = 1;i <= m;++i) c[i] += c[i - 1];
	for(int i = n;i >= 1;--i) sa[c[x[i]]--] = i;

	for(int k = 1;k <= n;k <<= 1) {
		int num = 0;
		for(int i = n - k + 1;i <= n;++i) y[++num] = i;
		for(int i = 1;i <= n;++i) if(sa[i] > k) y[++num] = sa[i] - k;
		for(int i = 1;i <= m;++i) c[i] = 0;
		for(int i = 1;i <= n;++i) ++c[x[i]];
		for(int i = 1;i <= m;++i) c[i] += c[i - 1];
		for(int i = n;i >= 1;--i) sa[c[x[y[i]]]--] = y[i];
		swap(x,y);
		num = 0;
		x[sa[1]] = ++num;
		for(int i = 2;i <= n;++i) {
			if(y[sa[i]] == y[sa[i - 1]] && y[sa[i] + k] == y[sa[i - 1] + k]) x[sa[i]] = num;
			else x[sa[i]] = ++num;
		}
		m = num;
		if(m == n) break;
	}
}
int rk[N];
void get_height() {
	for(int i = 1;i <= n;++i) rk[sa[i]] = i;
	int k = 0;
	for(int i = 1;i <= n;++i) {
		if(rk[i] == 1) continue;
		if(k) --k;
		int j = sa[rk[i] - 1];
		while(j + k <= n && i + k <= n && s[j + k] == s[i + k]) ++k;
		height[rk[i]] = k;
		// printf("%d %d %d
",i,j,k);
	}
}
int sta[N],top;
int main() {
	// freopen("P3804_1.in","r",stdin);
	scanf("%s",s + 1);
	n = strlen(s + 1);
	m = 'z';
	get_sa();

	get_height();
	// for(int i = 1;i <= n;++i) printf("%d ",height[i]);puts("");

	int ans = 0;
	for(int i = 1;i <= n + 1;++i) {
		while(top && height[i] < height[sta[top]]) {
			ans = max(ans,(i - sta[top - 1]) * height[sta[top]]);
			--top;
		}
		sta[++top] = i;
	}

/*
	for(int r = 1;r <= n;++r) {
		int mn = height[r];
		for(int l = r;l >= 1;--l) {
			mn = min(mn,height[l]);
			ans = max(ans,mn * (r - l + 2));
		}
	}*/

	cout<<ans;
	return 0;
}