POJ2774 Long Long Message
? 解题记录 ? ? 后缀自动机 ? ? 动态规划 ?    2017-10-01 11:19:28    577    0    0
Long Long Message
Time Limit: 4000MS Memory Limit: 131072K
Total Submissions: 31899 Accepted: 12873
Case Time Limit: 1000MS

Description

The little cat is majoring in physics in the capital of Byterland. A piece of sad news comes to him these days: his mother is getting ill. Being worried about spending so much on railway tickets (Byterland is such a big country, and he has to spend 16 shours on train to his hometown), he decided only to send SMS with his mother. 

The little cat lives in an unrich family, so he frequently comes to the mobile service center, to check how much money he has spent on SMS. Yesterday, the computer of service center was broken, and printed two very long messages. The brilliant little cat soon found out: 

1. All characters in messages are lowercase Latin letters, without punctuations and spaces. 
2. All SMS has been appended to each other – (i+1)-th SMS comes directly after the i-th one – that is why those two messages are quite long. 
3. His own SMS has been appended together, but possibly a great many redundancy characters appear leftwards and rightwards due to the broken computer. 
E.g: if his SMS is “motheriloveyou”, either long message printed by that machine, would possibly be one of “hahamotheriloveyou”, “motheriloveyoureally”, “motheriloveyouornot”, “bbbmotheriloveyouaaa”, etc. 
4. For these broken issues, the little cat has printed his original text twice (so there appears two very long messages). Even though the original text remains the same in two printed messages, the redundancy characters on both sides would be possibly different. 

You are given those two very long messages, and you have to output the length of the longest possible original text written by the little cat. 

Background: 
The SMS in Byterland mobile service are charging in dollars-per-byte. That is why the little cat is worrying about how long could the longest original text be. 

Why ask you to write a program? There are four resions: 
1. The little cat is so busy these days with physics lessons; 
2. The little cat wants to keep what he said to his mother seceret; 
3. POJ is such a great Online Judge; 
4. The little cat wants to earn some money from POJ, and try to persuade his mother to see the doctor :( 

Input

Two strings with lowercase letters on two of the input lines individually. Number of characters in each one will never exceed 100000.

Output

A single line with a single integer number – what is the maximum length of the original text written by the little cat.

Sample Input

yeshowmuchiloveyoumydearmotherreallyicannotbelieveit
yeaphowmuchiloveyoumydearmother

Sample Output

27

Source

POJ Monthly--2006.03.26,Zeyuan Zhu,"Dedicate to my great beloved mother."

复习了一下后缀自动机。虽然说学的时候很难,但是用起来就很便于理解又很好写了。一道公共子串,后缀自动机很巧妙的O(n)AC。十分经典。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn = 2e5 + 5;
int trie[maxn][26], len[maxn], fa[maxn], dp[maxn], last, cnt;
char s[maxn], ms[maxn];

void add(int x) {
	int id = s[x] - 'a';
	int np = ++cnt, p = last;
	last = np;
	len[np] = x + 1;
	while(p && !trie[p][id]) {
		trie[p][id] = np;
		p = fa[p];
	}
	if(p == 0) fa[np] = 1;
	else {
		int q = trie[p][id];
		if(len[p] + 1 == len[q]) fa[np] = q;
		else {
			int lca = ++cnt;
			len[lca] = len[p] + 1;
			fa[lca] = fa[q];
			fa[q] = fa[np] = lca;
			memcpy(trie[lca], trie[q], sizeof(trie[q]));
			while(p && trie[p][id] == q) {
				trie[p][id] = lca;
				p = fa[p];
			}
		}
	}
}

int work() {
	int p = 1, step = 0, ans = 0;
	int l = strlen(ms);
	for(register int i = 0; i < l; ++i) {
		int id = ms[i] - 'a';
		if(trie[p][id]) dp[trie[p][id]] = ++step, p = trie[p][id];
		else {
			while(p && !trie[p][id]) p = fa[p];
			if(p == 0) p = 1, step = 0;
			else step = dp[trie[p][id]] = len[p] + 1, p = trie[p][id];
		}
		ans = max(ans, dp[p]);
	}
	return ans;
}

int main() {
	last = cnt = 1;//从1是根节点
	scanf("%s", s);
	scanf("%s", ms);
	int l = strlen(s);
	for(register int i = 0; i < l; ++i) add(i);
	printf("%d", work());
	return 0;
}

 

上一篇: SPOJ1812 Longest Common Substring II

下一篇: POJ3259 Wormholes

577 人读过
立即登录, 发表评论.
没有帐号? 立即注册
0 条评论
文档导航