SP705 SUBST1 - New Distinct Substrings

\(\color{#0066ff}{ 题目描述 }\)

给定一个字符串，求该字符串含有的本质不同的子串数量。

\(\color{#0066ff}{输入格式}\)

T- number of test cases. T<=20; Each test case consists of one string, whose length is <= 50000

\(\color{#0066ff}{输出格式}\)

For each test case output one number saying the number of distinct substrings.

\(\color{#0066ff}{输入样例}\)

2
CCCCC
ABABA

\(\color{#0066ff}{输出样例}\)

5
9

\(\color{#0066ff}{数据范围与提示}\)

none

\(\color{#0066ff}{ 题解 }\)

本质不同字串？？？

这不就是自动机上所有节点维护的所有串吗

作为一个最简自动机，这才是真正的板子题吧qwq

\(ans = \sum{len[o] - len[fa[o]]}\)

#include<bits/stdc++.h>
using namespace std;
#define LL long long
LL in() {
	char ch; int x = 0, f = 1;
	while(!isdigit(ch = getchar()))(ch == '-') && (f = -f);
	for(x = ch ^ 48; isdigit(ch = getchar()); x = (x << 1) + (x << 3) + (ch ^ 48));
	return x * f;
}
const int maxn = 1e5 + 5;
struct SAM {
protected:
	struct node {
		node *ch[26], *fa;
		int len, siz;
		node(int len = 0, int siz = 0): fa(NULL), len(len), siz(siz) {
			memset(ch, 0, sizeof ch);
		}
	};
	node *root, *tail, *lst;
	node pool[maxn];
	node *extend(int c) {
		node *o = new(tail++) node(lst->len + 1, 1), *v = lst;
		for(; v && !v->ch[c]; v = v->fa) v->ch[c] = o;
		if(!v) o->fa = root;
		else if(v->len + 1 == v->ch[c]->len) o->fa = v->ch[c];
		else {
			node *n = new(tail++) node(v->len + 1), *d = v->ch[c];
			std::copy(d->ch, d->ch + 26, n->ch);
			n->fa = d->fa, d->fa = o->fa = n;
			for(; v && v->ch[c] == d; v = v->fa) v->ch[c] = n;
		}
		return lst = o;
	}
public:
	void clr() {
		tail = pool;
		root = lst = new(tail++) node();
	}
	SAM() { clr(); }
	LL ins(char *s) {
		LL ans = 0;
		for(char *p = s; *p; p++) {
			node *o = extend(*p - 'a');
			ans += o->len - o->fa->len;
		}
		return ans;
	}
}sam;
char s[maxn];
int main() {
	for(int T = in(); T --> 0;) {
		scanf("%s", s);
		printf("%lld\n", sam.ins(s));
		sam.clr();
	}
	return 0;
}