[SPOJ1812]Longest Common Substring II

[SPOJ1812]Longest Common Substring II

试题描述

A string is finite sequence of characters over a non-empty finite set Σ.

In this problem, Σ is the set of lowercase letters.

Substring, also called factor, is a consecutive sequence of characters occurrences at least once in a string.

Now your task is a bit harder, for some given strings, find the length of the longest common substring of them.

Here common substring means a substring of two or more strings.

输入

The input contains at most 10 lines, each line consists of no more than 100000 lowercase letters, representing a string.

输出

The length of the longest common substring. If such string doesn't exist, print "0" instead.

输入示例

alsdfkjfjkdsal
fdjskalajfkdsla
aaaajfaaaa

输出示例

2

数据规模及约定

见“输入”

题解

考虑把所有串连接起来构建后缀自动机（之前一看到这种题果断想后缀数组），那么我们在 parent 树上处理出每个节点子树中包含的串的集合。那么最后扫一遍每个节点，对于所有满的集合用 Max[i] 更新答案就好了。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <algorithm>
using namespace std;
 
int read() {
	int x = 0, f = 1; char c = getchar();
	while(!isdigit(c)){ if(c == '-') f = -1; c = getchar(); }
	while(isdigit(c)){ x = x * 10 + c - '0'; c = getchar(); }
	return x * f;
}
 
#define maxn 2000110
#define maxa 40
 
int rt, ToT, last, ch[maxn][maxa], par[maxn], Max[maxn], id[maxn];
 
char Str[maxn];
int S[maxn], Id[maxn], len;
 
void extend(int i) {
	int x = S[i], p = last, np = ++ToT; Max[np] = Max[p] + 1; id[np] = (Id[i] <= 10 ? (1 << Id[i]) : 0);
	last = np;
	while(p && !ch[p][x]) ch[p][x] = np, p = par[p];
	if(!p){ par[np] = rt; return ; }
	int q = ch[p][x];
	if(Max[q] == Max[p] + 1){ par[np] = q; return ; }
	int nq = ++ToT; Max[nq] = Max[p] + 1;
	memcpy(ch[nq], ch[q], sizeof(ch[q]));
	par[nq] = par[q];
	par[np] = par[q] = nq;
	while(p && ch[p][x] == q) ch[p][x] = nq, p = par[p];
	return ;
}
 
int sa[maxn], Ws[maxn];
void build() {
	for(int i = 1; i <= ToT; i++) Ws[len-Max[i]]++;
	for(int i = 1; i <= len; i++) Ws[i] += Ws[i-1];
	for(int i = ToT; i; i--) sa[Ws[len-Max[i]]--] = i;
	for(int i = 1; i <= ToT; i++) id[par[sa[i]]] |= id[sa[i]];
	return ;
}
 
int main() {
	rt = last = ToT = 1;
	int cnt = 0;
	while(~scanf("%s", Str)) {
		int n = strlen(Str);
		for(int i = 0; i < n; i++) S[++len] = Str[i] - 'a', Id[len] = cnt;
		S[++len] = cnt + 26; Id[len] = cnt + 11;
		cnt++;
	}
	
	for(int i = 1; i <= len; i++) extend(i);
	build();
	int all = (1 << cnt) - 1, ans = 0;
	for(int i = 1; i <= ToT; i++)
		if(id[i] == all) ans = max(ans, Max[i]);
	
	printf("%d
", ans);
	
	return 0;
}

然而 SPOJ 卡常数 T 飞~~~

考虑优化，因为这个最长公共子串一定出现在某一个字符串中，所以我们只需要对第一个串构造 SAM，对于后面的串，我们分别扫一遍在 SAM 上转移，然后再每个节点上更新一下当前匹配的最长长度信息，然后取取 min max 什么的就好了。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <algorithm>
using namespace std;

int read() {
	int x = 0, f = 1; char c = getchar();
	while(!isdigit(c)){ if(c == '-') f = -1; c = getchar(); }
	while(isdigit(c)){ x = x * 10 + c - '0'; c = getchar(); }
	return x * f;
}

#define maxn 200010
#define maxa 26

int rt, ToT, last, ch[maxn][maxa], par[maxn], Max[maxn], Ans[maxn], mn[maxn];

char Str[maxn];
int len;

void extend(int i) {
	int x = Str[i] - 'a', p = last, np = ++ToT; Max[np] = Max[p] + 1;
	last = np;
	while(p && !ch[p][x]) ch[p][x] = np, p = par[p];
	if(!p){ par[np] = rt; return ; }
	int q = ch[p][x];
	if(Max[q] == Max[p] + 1){ par[np] = q; return ; }
	int nq = ++ToT; Max[nq] = Max[p] + 1;
	memcpy(ch[nq], ch[q], sizeof(ch[q]));
	par[nq] = par[q];
	par[np] = par[q] = nq;
	while(p && ch[p][x] == q) ch[p][x] = nq, p = par[p];
	return ;
}
int sa[maxn], Ws[maxn];
void build() {
	for(int i = 1; i <= ToT; i++) Ws[len-Max[i]]++;
	for(int i = 1; i <= len; i++) Ws[i] += Ws[i-1];
	for(int i = ToT; i; i--) sa[Ws[len-Max[i]]--] = i;
	return ;
}
void get(int n) {
	int p = rt, tmp = 0;
	memset(Ans, 0, sizeof(Ans));
	for(int i = 0; i < n; i++) {
		int x = Str[i] - 'a';
		if(ch[p][x]) p = ch[p][x], tmp++;
		else {
			while(p && !ch[p][x]) p = par[p];
			if(!p) tmp = 0, p = 1;
			else tmp = Max[p] + 1, p = ch[p][x];
		}
		Ans[p] = max(Ans[p], tmp);
	}
	for(int i = 1; i <= ToT; i++) {
		int u = sa[i];
		mn[u] = min(mn[u], Ans[u]);
		if(par[u] && Ans[u]) Ans[par[u]] = Max[par[u]];
	}
	return ;
}

int main() {
	rt = last = ToT = 1;
	scanf("%s", Str); len = strlen(Str);
	for(int i = 0; i < len; i++) extend(i);
	build();
	for(int i = 1; i <= ToT; i++) mn[i] = Max[i];
	while(~scanf("%s", Str)) get(strlen(Str));
	
	int ans = 0;
	for(int i = 1; i <= ToT; i++) ans = max(ans, mn[i]);
	printf("%d
", ans);
	
	return 0;
}