POJ 2752 Seek the Name, Seek the Fame【EXKMP/KMP】

最新推荐文章于 2019-08-13 12:18:36 发布

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本文介绍了一种使用串匹配算法来找出字符串中既是前缀也是后缀的所有子串长度的方法。通过两个不同的实现方案，即EXKMP算法和KMP算法，文章详细展示了如何高效地解决这一问题。

Seek the Name, Seek the Fame

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 17408		Accepted: 8916

Description

The little cat is so famous, that many couples tramp over hill and dale to Byteland, and asked the little cat to give names to their newly-born babies. They seek the name, and at the same time seek the fame. In order to escape from such boring job, the innovative little cat works out an easy but fantastic algorithm:

Step1. Connect the father's name and the mother's name, to a new string S.
Step2. Find a proper prefix-suffix string of S (which is not only the prefix, but also the suffix of S).

Example: Father='ala', Mother='la', we have S = 'ala'+'la' = 'alala'. Potential prefix-suffix strings of S are {'a', 'ala', 'alala'}. Given the string S, could you help the little cat to write a program to calculate the length of possible prefix-suffix strings of S? (He might thank you by giving your baby a name:)

Input

The input contains a number of test cases. Each test case occupies a single line that contains the string S described above.

Restrictions: Only lowercase letters may appear in the input. 1 <= Length of S <= 400000.

Output

For each test case, output a single line with integer numbers in increasing order, denoting the possible length of the new baby's name.

Sample Input

ababcababababcabab
aaaaa

Sample Output

2 4 9 18
1 2 3 4 5

题意：求一个串的前i位是否等于后i位递增输出所有i+1；

思路：用EXKMP求出ene[i]是否等于len-i ；

失误：题看了好久才看懂，不理解说的前缀和后缀是什么；

AC代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

const int MAXN=1e6+55;
char str[MAXN]; int ene[MAXN];

void Get_ene(char *T,int *ene,int L)
{
	ene[0]=L; int i=0;
	while(i+1<L&&T[i]==T[i+1]) ++i;
	ene[1]=i; int id=1;
	for(i=2;i<L;++i)
	{
		int mx=id+ene[id]-1,ll=ene[i-id];
		if(i+ll-1>=mx)
		{
			int j=max(0,mx-i+1);
			while(i+j<L&&T[j]==T[i+j]) ++j;
			ene[i]=j; id=i;
		}
		else ene[i]=ll;
	 } 
}
int main()
{
    int i;
    while(~scanf("%s",str))
    {
    	int len=strlen(str);
    	Get_ene(str,ene,len);
    	int cnt=0;
    	for(i=len-1;i>=0;--i)
    	{
    		if(ene[i]==len-i){
				if(cnt) printf(" ");
    			if(!cnt) ++cnt;	
    			printf("%d",ene[i]); 
			} 
		}
		printf("\n");
	}
	return 0;
 }

KMP的看别人的思路（值得思考还是next的含义）：

#include<cstdio>
#include<cstring>
#include<stack>
using namespace std;

const int MAXN=1e6+33;
char str[MAXN]; int ne[MAXN];

void Get_ne(char *T,int *ne,int L)
{
	ne[0]=-1; int i=0,j=-1;
	while(i<L)
	{
		if(j==-1||T[i]==T[j]) ne[++i]=++j;
		else j=ne[j];
	}
}
int main()
{
	int i;
	while(~scanf("%s",str))
	{
		int len=strlen(str);
		Get_ne(str,ne,len);
		int tem=len; stack<int> STA;
		while(tem)
		{
			STA.push(tem); 
			tem=ne[tem];
	    }
	    bool  falg=false;
	    while(!STA.empty()) 
	    {
	    	if(falg) printf(" ");
	    	if(!falg) falg=true;
	    	printf("%d",STA.top());
			STA.pop(); 
		}
		printf("\n");
	}
	return 0;
 }