poj2774:后缀数组之殇

代码

不知道用这样的标题合不合适..总而言之,在我被后缀数组折磨了十余天后,我终于掌握了一种非主流的做法:O(nlog^2 n)构造法..在此我对在《高级数据结构》中介绍的后缀数组构造代码有很深的疑问..因为我发现我对着模板打出来的程序根本无法算出正确的后缀数组..晕

于是我使用的是《挑战程序设计竞赛》中的O(nlog^2 n)模板,真的很好理解,打算在暑假介绍给我亲爱的师弟们..

折腾了这么久,总算有点成效了!于是赶紧来A一题~

Long Long Message
Time Limit: 4000MS Memory Limit: 131072K
Total Submissions: 26259 Accepted: 10687
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."
题意:求两个字符串的LCS(Longest Common Substring,避免歧义)。
做法:
合并为一个字符串,用特殊的符号隔开(我之前用r隔开都能AC..Orz),然后求height数组(LCS数组)的最大值即可,注意判断是否来自不同的字符串。
为什么最长字串一定来自排名相邻的后缀?这个已经被证明了,有兴趣的读者可以上网查阅资料,这里不作证明。
代码如下:
/*Source Code

Problem: 2774		User: aclolicon
Memory: 3912K		Time: 1907MS
Language: G++		Result: Accepted
Source Code*/
#include
#include
#include 
#include
#define MAXN 200050
using namespace std;
string s, t;
int n, m, k;
int rank[MAXN], sa[MAXN], height[MAXN]; 
int tmpsa[MAXN];
bool cmp(int i, int j){
	if (rank[i] != rank[j]) return rank[i] < rank[j];
	int ri = i + k <= n? rank[i + k]: -1;
	int rj = j + k <= n? rank[j + k]: -1;
	return ri < rj;
}
void buildSA(){
	int n = s.length(); 
	int tmp[MAXN];
	for (int i = 0; i <= n; i++){
		sa[i] = i;
		rank[i] = i < n? s[i]: -1;
	}
	
	for (k = 1; k <= n; k *= 2){
		sort(sa, sa + n + 1, cmp);
		
		tmp[sa[0]] = 0;
		for (int i = 1; i <= n; i++){
			tmp[sa[i]] = tmp[sa[i - 1]] + (cmp(sa[i - 1], sa[i])? 1: 0);
		}
		for (int i = 0; i <= n; i++){
			rank[i] = tmp[i];
		}
	}
}

void callheight(){
	int n = s.length(); 
	for (int i = 0; i <= n; i++) rank[sa[i]] = i;
	int h = 0;
	height[0] = 0;
	for (int i = 0; i < n; i++){
		int j = sa[rank[i] - 1];
		if (h > 0) h--;
		for (; j + h < n && i + h < n; h++){
			if (s[j + h] != s[i + h]) break;
		} 
		height[rank[i] - 1] = h;
	}
}

int main(){
//	freopen("c.in", "r", stdin);
	int len;
	while(cin >> s){
		cin >> t;
		len = s.length();
		s += 'r' + t;	
		n = s.length();	
		buildSA();
		callheight();
		int maxh = -1;
//		for (int i = 0; i < n; i++) cout << sa[i] << " " ;
		for (int i = 0; i < s.length(); i++){
			if((sa[i] < len) != (sa[i + 1] < len)){
				maxh = max(maxh, height[i]);
			}
		}
		cout << maxh << endl;
	}

}

之前一直WA到吐血..今天终于搞定一道水题了,Orz,下一次要向1/8个男人进发。

我还没有学会写个人说明!

2 条评论

  1. WDWD,我发了一文章~~看看呗

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