KMP算法模板
例题1链接：点击打开链接
原题：
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 24044 | Accepted: 9633 |
Description
The French author Georges Perec (1936–1982) once wrote a book, La disparition, without the letter 'e'. He was a member of the Oulipo group. A quote from the book:
Tout avait Pair normal, mais tout s’affirmait faux. Tout avait Fair normal, d’abord, puis surgissait l’inhumain, l’affolant. Il aurait voulu savoir où s’articulait l’association qui l’unissait au roman : stir son tapis, assaillant à tout instant son imagination, l’intuition d’un tabou, la vision d’un mal obscur, d’un quoi vacant, d’un non-dit : la vision, l’avision d’un oubli commandant tout, où s’abolissait la raison : tout avait l’air normal mais…
Perec would probably have scored high (or rather, low) in the following contest. People are asked to write a perhaps even meaningful text on some subject with as few occurrences of a given “word” as possible. Our task is to provide the jury with a program that counts these occurrences, in order to obtain a ranking of the competitors. These competitors often write very long texts with nonsense meaning; a sequence of 500,000 consecutive 'T's is not unusual. And they never use spaces.
So we want to quickly find out how often a word, i.e., a given string, occurs in a text. More formally: given the alphabet {'A', 'B', 'C', …, 'Z'} and two finite strings over that alphabet, a word W and a textT, count the number of occurrences of W in T. All the consecutive characters of W must exactly match consecutive characters of T. Occurrences may overlap.
Input
The first line of the input file contains a single number: the number of test cases to follow. Each test case has the following format:
- One line with the word W, a string over {'A', 'B', 'C', …, 'Z'}, with 1 ≤ |W| ≤ 10,000 (here |W| denotes the length of the string W).
- One line with the text T, a string over {'A', 'B', 'C', …, 'Z'}, with |W| ≤ |T| ≤ 1,000,000.
Output
For every test case in the input file, the output should contain a single number, on a single line: the number of occurrences of the word W in the text T.
Sample Input
3 BAPC BAPC AZA AZAZAZA VERDI AVERDXIVYERDIAN
Sample Output
1 3 0
Source
AC code：
#include<stdio.h>
#include<string.h>char p[1111111],t[1111111];
int next[1111111],lenp,lent;
void get()
{
int i,j;
i=0;
next[i]=-1;
j=-1;
while(i<=lenp)
{
if(j==-1||p[i]==p[j])
{
i++;
j++;
next[i]=j;
}
else
j=next[j];
}
}
int main()
{
int i,j,cas,cnt;
while(scanf("%d",&cas)!=EOF)
{
while(cas--)
{
scanf("%s",p);
scanf("%s",t);
lenp=strlen(p);
lent=strlen(t);
get();
i=-1;
j=-1;
cnt=0;
while(i<=lent)
{
if(j==-1||p[j]==t[i])
{
i++;
j++;
}
else
j=next[j];
if(j==lenp)
{
j--;
i--;
j=next[j];
cnt++;
}
}
printf("%d\n",cnt);
}
}
return 0;
}
提取模板：
void Get_next()
{
int i=0,j=-1;
next[0]=-1;
while(i<=Tlen)
{
if(j==-1||chT[i]==chT[j])
{
i++; j++;
next[i]=j;
}
else
j=next[j];
}
}
int Kmp(int u)
{
int i=u-1,j=-1,m=0; //字符串从0开始存储的
while(i<=TTlen)
{
if(j==-1||ch[i]==chT[j])
{
i++; j++;
}
else
j=next[j];
if(j>=Tlen) //**注意这里要j--,i++,因为上面匹配成功了i和j的值加多了1
{
m++;
j--; i--;
j=next[j]; //返回
}
}
return m;
}
例题2链接：点击打开链接
原题：
Theme Section
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1239 Accepted Submission(s): 646
To get well prepared for the festival, the hosts want to know the maximum possible length of the theme section of each song. Can you help us?
5 xy abc aaa aaaaba aaxoaaaaa
0 0 1 1 2
AC code：
#include<stdio.h>
#include<string.h>
char s[1111111];
int next[1111111],len,ans[1111111];
void get()
{
int i,j;
i=0;
next[0]=-1;
j=-1;
while(i<=len)
{
if(j==-1||s[j]==s[i])
{
i++;
j++;
next[i]=j;
}
else
j=next[j];
}
}
int main()
{
int i,j,k,cas,fg;
while(scanf("%d",&cas)!=EOF)
{
while(cas--)
{
scanf("%s",s);
len=strlen(s);
get();
ans[0]=len;
i=0;j=len;
while(next[j]>0)
{
ans[++i]=next[j];
j=next[j];
}
fg=0;
for(j=0;j<=i;j++)
{
if(ans[j]>len/3)
continue;
else
{
for(k=2*ans[j];k<=len-ans[j];k++)
{
if(next[k]==ans[j])
{
fg=1;
printf("%d\n",ans[j]);
break;
}
}
}
if(fg)
break;
}
if(fg==0)
printf("0\n");
}
}
return 0;
}
提取模板：
void Get_next() //求字符串的next[]前缀数组
{
int i=0,j=-1;
next[0]=-1;
while(i<=Tlen)
{
if(j==-1||ch[i]==ch[j])
{
i++; j++;
next[i]=j;
}
else
j=next[j];
}
}
例题3链接：点击打开链接
原题：
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 12324 | Accepted: 5355 |
Description
As an IBM researcher, you have been tasked with writing a program that will find commonalities amongst given snippets of DNA that can be correlated with individual survey information to identify new genetic markers.
A DNA base sequence is noted by listing the nitrogen bases in the order in which they are found in the molecule. There are four bases: adenine (A), thymine (T), guanine (G), and cytosine (C). A 6-base DNA sequence could be represented as TAGACC.
Given a set of DNA base sequences, determine the longest series of bases that occurs in all of the sequences.
Input
- A single positive integer m (2 <= m <= 10) indicating the number of base sequences in this dataset.
- m lines each containing a single base sequence consisting of 60 bases.
Output
Sample Input
3 2 GATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA 3 GATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATA GATACTAGATACTAGATACTAGATACTAAAGGAAAGGGAAAAGGGGAAAAAGGGGGAAAA GATACCAGATACCAGATACCAGATACCAAAGGAAAGGGAAAAGGGGAAAAAGGGGGAAAA 3 CATCATCATCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC ACATCATCATAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AACATCATCATTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTT
Sample Output
no significant commonalities AGATAC CATCATCAT
Source
#include<string.h>
char s[11][70],t[70],t1[70];
int next[70],len,cnt;
void get_next(char p[],int lenp)
{
int i,j;
i=0;
next[0]=-1;
j=-1;
while(i<=lenp)
{
if(j==-1||p[i]==p[j])
{
i++;
j++;
next[i]=j;
}
else
j=next[j];
}
}
int kmp(char s[],int lens,char p[],int lenp)
{
get_next(p,lenp);
int i,j,start;
start=-1;
i=-1;
j=-1;
while(i<=lens)
{
if(j==-1||p[j]==s[i])
{
i++;
j++;
}
else
j=next[j];
if(j==lenp)
{
start=i-lenp+1;//求出在目标字符串中从第几个开始就完全匹配，找不到为-1
break;
}
}
return start;
}
int main()
{
int cas,n,i,j,k,m,c,st,maxlen;
while(scanf("%d",&cas)!=EOF)
{
while(cas--)
{
scanf("%d",&n);
for(i=1;i<=n;i++)
{
scanf("%s",s[i]);
}
len=strlen(s[1]);
maxlen=0;
for(i=0;i<len-1;i++)
{
for(c=1;c<=len-1-i;c++)
{
j=len-c;
if(j-i+1<maxlen)
break;
cnt=0;
for(m=i;m<=j;m++)
{
t[cnt]=s[1][m];
cnt++;
}
t[cnt]='\0';
for(k=2;k<=n;k++)
{
st=kmp(s[k],len,t,cnt);
if(st==-1)
break;
}
if(k==n+1)
{
if(cnt==maxlen)
{
if(strcmp(t,t1)<0)
strcpy(t1,t);
}
if(cnt>maxlen)
{
maxlen=cnt;
strcpy(t1,t);
}
}
}
}
if(maxlen<3)
printf("no significant commonalities\n");
else
puts(t1);
}
}
return 0;
}
{
int i,j;
i=0;
next[0]=-1;
j=-1;
while(i<=lenp)
{
if(j==-1||p[i]==p[j])
{
i++;
j++;
next[i]=j;
}
else
j=next[j];
}
}
int kmp(char s[],int lens,char p[],int lenp)
{
get_next(p,lenp);
int i,j,start;
start=-1;
i=-1;
j=-1;
while(i<=lens)
{
if(j==-1||p[j]==s[i])
{
i++;
j++;
}
else
j=next[j];
if(j==lenp)
{
start=i-lenp+1;//求出在目标字符串中从第几个开始就完全匹配，找不到为-1
break;
}
}
return start;
}
例题4链接：点击打开链接
原题：
| Time Limit: 3000MS | Memory Limit: 65536K | |
| Total Submissions: 5078 | Accepted: 1885 |
Description
Beside other services, ACM helps companies to clearly state their “corporate identity”, which includes company logo but also other signs, like trademarks. One of such companies is Internet Building Masters (IBM), which has recently asked ACM for a help with their new identity. IBM do not want to change their existing logos and trademarks completely, because their customers are used to the old ones. Therefore, ACM will only change existing trademarks instead of creating new ones.
After several other proposals, it was decided to take all existing trademarks and find the longest common sequence of letters that is contained in all of them. This sequence will be graphically emphasized to form a new logo. Then, the old trademarks may still be used while showing the new identity.
Your task is to find such a sequence.
Input
The input contains several tasks. Each task begins with a line containing a positive integer N, the number of trademarks (2 ≤ N ≤ 4000). The number is followed by N lines, each containing one trademark. Trademarks will be composed only from lowercase letters, the length of each trademark will be at least 1 and at most 200 characters.
After the last trademark, the next task begins. The last task is followed by a line containing zero.
Output
For each task, output a single line containing the longest string contained as a substring in all trademarks. If there are several strings of the same length, print the one that is lexicographically smallest. If there is no such non-empty string, output the words “IDENTITY LOST” instead.
Sample Input
3 aabbaabb abbababb bbbbbabb 2 xyz abc 0
Sample Output
abb IDENTITY LOST
Source
AC code：
#include<stdio.h>
#include<string.h>
char s[4001][201],s1[201],ans[201];
int next[201],lens[4001];
void get(char p[],int lenp)
{
int i,j;
i=0;
next[0]=-1;
j=-1;
while(i<=lenp)
{
if(j==-1||p[i]==p[j])
{
i++;
j++;
next[i]=j;
}
else
j=next[j];
}
}
int kmp(char t[],int lent,char p[],int lenp)
{
get(p,lenp);
int i,j,start;
i=-1;j=-1;start=-1;
while(i<=lent)
{
if(j==-1||p[j]==t[i])
{
i++;
j++;
}
else
j=next[j];
if(j==lenp)
{
start=i-lenp+1;
break;
}
}
return start;
}
int main()
{
int n,i,j,k,m,minlen,mini,cnt,st,maxlen;
while(scanf("%d",&n)!=EOF&&n)
{
minlen=300;
for(i=1;i<=n;i++)
{
scanf("%s",&s[i]);
lens[i]=strlen(s[i]);
if(minlen>lens[i])
{
minlen=lens[i];
mini=i;
}
}
maxlen=0;
for(i=0;i<=minlen-2;i++)
{
for(j=1;j<=minlen-i;j++)
{
if(minlen-j-i+1<maxlen)
break;
cnt=0;
for(k=i;k<=minlen-j;k++)
{
s1[cnt++]=s[mini][k];
}
s1[cnt]='\0';
for(m=1;m<=n;m++)
{
if(m==mini)
continue;
st=kmp(s[m],lens[m],s1,cnt);
if(st==-1)
break;
}
if(m==n+1)
{
if(maxlen==cnt)
{
if(strcmp(ans,s1)>0)
strcpy(ans,s1);
}
if(maxlen<cnt)
{
maxlen=cnt;
strcpy(ans,s1);
}
}
}
}
if(maxlen==0)
printf("IDENTITY LOST\n");
else
puts(ans);
}
return 0;
}
例题5链接：点击打开链接
原题：
| Time Limit: 3000MS | Memory Limit: 65536K | |
| Total Submissions: 32616 | Accepted: 13602 |
Description
Input
Output
Sample Input
abcd aaaa ababab .
Sample Output
1 4 3
Hint
Source
该题代码和分析转自网址：http://blog.csdn.net/qq7366020/article/details/12968747
题目大意: 给出一个由某个串重复有限次得到的字符串
求重复次数最多是多少，既找出最小重复子串
解题思路: 字符串abcabcabc的next[ ]值为
0 1 2 3 4 5 6 7 8 9
a b c a b c a b c
-1 0 0 0 1 2 3 4 5 6
假设主串为abcabcabcX，模式串为abcabcabcK，Kmp的匹配过程
匹配到第10个字符发现不符:
a b c a b c a b c X
a b c a b c a b c K
不符合，j=next[j]
a b c a b c a b c X
a b c a b c a b c K
不符合，j=next[j]
a b c a b c a b c X
a b c a b c a b c K;
由next数组的定义可得： S[0-2]=S[3-5]=S[6-8]
当且仅当len%(len-next[len])=0时,len-next[len]是字符串的最小循环节
代码:
- #include <stdio.h>
- #include <string.h>
- #define MAX 1000005
- char ch[MAX];
- int next[MAX];
- int Get_next(int Tlen)
- {
- int i=0,j=-1;
- next[0]=-1;
- while(i<Tlen)
- {
- if(j==-1||ch[i]==ch[j])
- {
- i++; j++;
- next[i]=j;
- }
- else
- j=next[j];
- }
- i=Tlen-j; //**
- if(Tlen%i==0) //**
- return Tlen/i; //**
- return 1;
- }
- int main()
- {
- int Tlen;
- while(scanf("%s",ch)!=EOF)
- {
- if(ch[0]=='.')
- break;
- Tlen=strlen(ch);
- printf("%d\n",Get_next(Tlen));
- }
- return 0;
- }
原题：
| Time Limit: 3000MS | Memory Limit: 30000K | |
| Total Submissions: 13365 | Accepted: 6296 |
Description
Input
number zero on it.
Output
Sample Input
3 aaa 12 aabaabaabaab 0
Sample Output
Test case #1 2 2 3 3 Test case #2 2 2 6 2 9 3 12 4
Source
code：
#include<stdio.h>
#include<string.h>
char s[1000111];
int next[1000111],ans[1000111];
void get(char p[],int lenp)
{
int i,j;
i=0;
j=-1;
next[0]=-1;
while(i<=lenp)
{
if(j==-1||p[j]==p[i])
{
i++;
j++;
next[i]=j;
}
else
j=next[j];
}
for(i=1;i<=lenp-1;i++)
{
if((i+1)%(i+1-next[i+1])==0&&(i+1)!=(i+1-next[i+1]))//由于不输出1
printf("%d %d\n",i+1,(i+1)/(i+1-next[i+1]));
}
}
int main()
{
int len,n,cas;
cas=0;
while(scanf("%d",&n)!=EOF&&n)
{
cas++;
scanf("%s",s);
len=strlen(s);
printf("Test case #%d\n",cas);
get(s,len);
printf("\n");
}
return 0;
}
例题7链接：http://poj.org/problem?id=2752
原题：
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 12266 | Accepted: 6032 |
Description
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
Restrictions: Only lowercase letters may appear in the input. 1 <= Length of S <= 400000.
Output
Sample Input
ababcababababcabab aaaaa
Sample Output
2 4 9 18 1 2 3 4 5
Source
大意：
给出一个字符串A,求A有多少个前缀同时也是后缀，从小到大输出这些前缀的长度。
分析：KMP
对于长度为len的字符串，由next的定义知：
A[0]A[1]...A[next[len]-1]=A[len-next[len]]...A[len-1]此时A[0]A[1]...A[next[len]-1]为一个符合条件的前缀
有A[0]A[1]....A[next[next[len]]-1] = A[len-next[next[len] - next[next[len]]]...A[next[len]-1],故A[0]A[1]....A[next[next[len]]-1]也是一个符合条件的前缀
故从len=>next[len]=>next[next[len]] ....=>直到某个next[]为0均为合法答案，注意当首位单词相同时，也为答案。
code：
#include<string.h>
#include<stdio.h>
char s[400111];
int next[400111],ans[400111],len;
void get(char p[],int lenp)
{
int i,j;
i=0;j=-1;
next[0]=-1;
while(i<=lenp)
{
if(j==-1||p[j]==p[i])
{
i++;j++;
next[i]=j;
}
else
j=next[j];
}
}
int main()
{
int n,cnt,i;
while(scanf("%s",s)!=EOF)
{
len=strlen(s);
get(s,len);
ans[1]=len;
cnt=1;
n=next[len];
while(n)
{
cnt++;
ans[cnt]=n;
n=next[n];
}
for(i=cnt;i>=1;i--)
printf("%d ",ans[i]);
printf("\n");
}
return 0;
}
解题思路分析亦可参考网址：http://www.doc88.com/p-997525236064.html
例题8链接：点击打开链接
原题：
Cyclic Nacklace
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2870 Accepted Submission(s): 1324
As Christmas is around the corner, Boys are busy in choosing christmas presents to send to their girlfriends. It is believed that chain bracelet is a good choice. However, Things are not always so simple, as is known to everyone, girl's fond of the colorful decoration to make bracelet appears vivid and lively, meanwhile they want to display their mature side as college students. after CC understands the girls demands, he intends to sell the chain bracelet called CharmBracelet. The CharmBracelet is made up with colorful pearls to show girls' lively, and the most important thing is that it must be connected by a cyclic chain which means the color of pearls are cyclic connected from the left to right. And the cyclic count must be more than one. If you connect the leftmost pearl and the rightmost pearl of such chain, you can make a CharmBracelet. Just like the pictrue below, this CharmBracelet's cycle is 9 and its cyclic count is 2:
Now CC has brought in some ordinary bracelet chains, he wants to buy minimum number of pearls to make CharmBracelets so that he can save more money. but when remaking the bracelet, he can only add color pearls to the left end and right end of the chain, that is to say, adding to the middle is forbidden.
CC is satisfied with his ideas and ask you for help.
Each test case contains only one line describe the original ordinary chain to be remade. Each character in the string stands for one pearl and there are 26 kinds of pearls being described by 'a' ~'z' characters. The length of the string Len: ( 3 <= Len <= 100000 ).
3 aaa abca abcde
0 2 5
code：
#include<stdio.h>
#include<string.h>
char s[100111];
int next[100111],len;
void get(char p[],int lenp)
{
int i,j;
i=0;
j=-1;
next[0]=-1;
while(i<=lenp)
{
if(j==-1||p[j]==p[i])
{
i++;
j++;
next[i]=j;
}
else
j=next[j];
}
}
int main()
{
int n,i,j;
while(scanf("%d",&n)!=EOF)
{
while(n--)
{
scanf("%s",s);
len=strlen(s);
get(s,len);
if(len%(len-next[len])==0)
{
if(next[len]==0)//相当于len/(len-next[len])==1的情况
printf("%d\n",len);
else
printf("0\n");
}
else
{
i=len%(len-next[len])-1;
j=len-1;
while(i>=0)
{
if(s[i]==s[j])
{
i--;
j--;
}
else
break;
}
if(i==-1)
printf("%d\n",len-next[len]-len%(len-next[len]));
else
printf("%d\n",len);
}
}
}
return 0;
}
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题目反思与总结：
当所给字符串中无循环节时，如abcdeab ， 我们求出其next数组的对应值（注意按照标准的next数组要多求一位，计算时向后推一位计算，因为标准的next数组第n个数表示的是第n-1个字符的匹配情况），然后用字符串长度L-next[L]即可，此时上例中显然next[L]为2，即为多出的ab，所以循环节就是7-2=5；
当所给字符串中有多个字符串时，如abxabxabxab,此时next[L]为8,11-8=3，亦是循环节，我们可以看出L-next[L]就是循环节，其实在一个多个循环节的字符串中，除了第一个循环节中字符外，其他每个位置的下标n-next[n]都是循环节（第一个循环节中的最后一个位置也可）。
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