KMP算法的next、next value数组代码实现及POJ3461
昨天中午弄懂了数组的手工计算方法之后，根据书上例题解出了一道KMP算法的匹配题。
我用了next 和nextval两种解决方法，其实就是数组实现的代码片不同。
w表示给定的模式字符串
next数组代码实现如下：
int next[maxw],j=0,i;
next[0]=-1;
next[1]=0;
for(i=2; i<=strlen(w); ++i)
{
while(j>=0&&w[j]!=w[i-1])
j=next[j];
next[i]=++j;
}
nextval数组代码实现如下：
int nextval[maxw],i=0,j=-1;
nextval[0]=-1;
while(i下面贴一道POJ上的例题：
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 29204 | Accepted: 11704 |
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 text T, 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
这道题就是求匹配过程中子串在主串中出现了多少次。
next版：
#include
#include
#include
using namespace std;
const int maxw=10000+10;
const int maxt=1000000+10;
int match(char w[],char s[],int next [])
{
int cnt=0,p=0,cur=0,slen,wlen;
slen=strlen(s);
wlen=strlen(w);
while(cur=0)
{
p=next[p];
}
else
{
++cur;
p=0;
}
if(p==wlen)
{
++cnt;
p=next[p];
}
}
return cnt;
}
int main()
{
int loop;
scanf("%d",&loop);
while(loop--)
{
char w[maxw],t[maxt];
scanf("%s%s",w,t);
int next[maxw],p=0,cur;
next[0]=-1;
next[1]=0;
for(cur=2; cur<=strlen(w); ++cur)
{
while(p>=0&&w[p]!=w[cur-1])
p=next[p];
next[cur]=++p;
}
printf("%d\n",match(w,t,next));
}
return 0;
} nextval版：
#include
#include
#include
using namespace std;
const int maxw=10000+10;
const int maxt=1000000+10;
int match(char w[],char s[],int next [])
{
int cnt=0,p=0,i=0,slen,wlen;
slen=strlen(s);
wlen=strlen(w);
while(i=0)
{
p=next[p];
}
else
{
++i;
p=0;
}
if(p==wlen)
{
++cnt;
p=next[p];
}
}
return cnt;
}
int main()
{
int loop;
scanf("%d",&loop);
while(loop--)
{
char w[maxw],t[maxt];
scanf("%s%s",w,t);
int nextval[maxw],i,j;
i=0;
nextval[0]=-1;
j=-1;
while(i 比较内存占用和运行时长，发现就这道题而言，nextval的用时要少但是内存占用较多。
总之，nextval是对next优化改进后的方法，效率会提高。
我对KMP算法的初步学习大概就是这么多认识，当然是木有BF算法那么好理解，但是KMP又快又好用阿~~
希望我能就这么坚持下去吧，即使脑子木有人家那么灵活但是如果多花花时间能弄出来我也是挺开心哒~~