poj 2752 Seek the Name, Seek the Fame【KMP】
來源:程序員人生 發布時間:2014-12-09 08:24:31 閱讀次數:2997次
點擊打開題目
Seek the Name, Seek the Fame
Time Limit: 2000MS |
|
Memory Limit: 65536K |
Total Submissions: 12651 |
|
Accepted: 6214 |
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
大致題意:
給出1個字符串str,求出str中存在多少子串,使得這些子串既是str的前綴,又是str的后綴。從小到大順次輸出這些子串的長度
解題思路:重在理解KMP算法中next數組的含義,遞歸求解
#include<cstdio>
#define maxn 400002
char str[maxn];
int next[maxn],len,s;
void GetNext(){
int i=0,j=⑴;
next[0]=⑴;
while(str[i]){
if(j==⑴||str[i]==str[j]){
++i;++j;next[i]=j;
}else j=next[j];
}
len=i;
}
void GetVal(int n){
if(next[n]==0) return ;
GetVal(next[n]);
printf("%d ",next[n]);
}
int main(){
while(scanf("%s",str)==1){
GetNext();
GetVal(len);
printf("%d
",len);
}return 0;
}
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