Seek the Name, Seek the Fame
Problem's Link: http://poj.org/problem?id=2752
Mean:
给你一个字符串，求这个字符串中有多少前缀是和后缀相等的， 按递增顺序输出这些前缀的长度。
analyse:
KMP之next数组的运用。
next[k]表示的是s[0....next[k]] 和s[k-next[k] .... k] 这段是相等的，即以k结尾的最长相等前后缀长度。
next[k-1]表示的是与s[0....k]具有最长后缀，且与s[0...k]具有最长前缀的位置。也就是在k位置失配后需要移到next[k-1]的位置。
这里特别说明一下，失配后移到next[k]还是next[k-1]，取决于next数组的开始编号。如果从0开始编号，那就移到next[k-1];从1开始编号，就移到next[k]。
Time complexity: O(N)
Source code:
/*
* this code is made by crazyacking
* Verdict: Accepted
* Submission Date: 2015-07-28-08.55
* Time: 0MS
* Memory: 137KB
*/
#include <queue>
#include <cstdio>
#include <set>
#include <string>
#include <stack>
#include <cmath>
#include <climits>
#include <map>
#include <cstdlib>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
#define LL long long
#define ULL unsigned long long
using namespace std;
const int MAXN = 400010;
int n;
char s [ MAXN ];
int Next [ MAXN ];
int ans [ MAXN ];
void getNext()
{
Next [ 0 ] = 0;
for( int i = 1 , k = 0; i <n; ++ i)
{
while(s [ i ] !=s [ k ] && k) k = Next [ k - 1 ];
if(s [ i ] ==s [ k ]) ++ k;
Next [ i ] = k;
}
}
int main()
{
ios_base :: sync_with_stdio( false);
cin . tie( 0);
while( ~ scanf( "%s" ,s))
{
n = strlen(s);
getNext();
int k =n - 1;
int cnt = 0;
while( Next [ k ])
{
ans [ cnt ++ ] = Next [ k ];
k = Next [ k - 1 ];
}
ans [ cnt ++ ] =n;
sort( ans , ans + cnt);
for( int i = 0; i < cnt; ++ i)
{
printf( i == cnt - 1 ? "%d \n " : "%d " , ans [ i ]);
}
}
return 0;
}
/*
*/
* this code is made by crazyacking
* Verdict: Accepted
* Submission Date: 2015-07-28-08.55
* Time: 0MS
* Memory: 137KB
*/
#include <queue>
#include <cstdio>
#include <set>
#include <string>
#include <stack>
#include <cmath>
#include <climits>
#include <map>
#include <cstdlib>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
#define LL long long
#define ULL unsigned long long
using namespace std;
const int MAXN = 400010;
int n;
char s [ MAXN ];
int Next [ MAXN ];
int ans [ MAXN ];
void getNext()
{
Next [ 0 ] = 0;
for( int i = 1 , k = 0; i <n; ++ i)
{
while(s [ i ] !=s [ k ] && k) k = Next [ k - 1 ];
if(s [ i ] ==s [ k ]) ++ k;
Next [ i ] = k;
}
}
int main()
{
ios_base :: sync_with_stdio( false);
cin . tie( 0);
while( ~ scanf( "%s" ,s))
{
n = strlen(s);
getNext();
int k =n - 1;
int cnt = 0;
while( Next [ k ])
{
ans [ cnt ++ ] = Next [ k ];
k = Next [ k - 1 ];
}
ans [ cnt ++ ] =n;
sort( ans , ans + cnt);
for( int i = 0; i < cnt; ++ i)
{
printf( i == cnt - 1 ? "%d \n " : "%d " , ans [ i ]);
}
}
return 0;
}
/*
*/
