Seek the Name, Seek the Fame
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 14188 | Accepted: 7068 |
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：给一个字符串S，判断在什么下标的时候，前缀和后缀相等,输出前缀和后缀相等的点。
分析：next数组的一种很巧妙的用法
next数组表示的意义是当前下标前面k字符和开头的前面k个字符相等
所以就会有xy=ab(用xy表示x~y的这一段)，则next[b]=y，那么下次就从y这个位置开始匹配
如果xk=wy，因为xy=ab，故wy=lb，所以xk=lb，就得到了前缀和后缀相等。
如此匹配直到next[i]=-1，期间记录下下标
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<string>
#include<iostream>
#include<cstring>
#include<cmath>
#include<stack>
#include<queue>
#include<vector>
#include<map>
#include<stdlib.h>
#include<algorithm>
#define LL __int64
using namespace std;
const int MAXN=100000+5;
char str[MAXN];
int nex[MAXN];
int ans[MAXN];
int n;
void getnext()
{
int j=0,k=-1;
nex[0]=-1;
while(j<n)
{
if(k==-1 || str[j]==str[k])
nex[++j]=++k;
else
k=nex[k];
}
}
int main()
{
//freopen("in.txt","r",stdin);
while(scanf("%s",str)!=EOF)
{
n=strlen(str);
getnext();
int i=n,tot=0;
while(nex[i]!=-1)
{
ans[tot++]=i;
i=nex[i];
}
for(int i=tot-1;i>=0;i--)
{
if(i==tot-1) printf("%d",ans[i]);
else printf(" %d",ans[i]);
}
printf("\n");
}
return 0;
}