KMP算法详解

 

http://www.matrix67.com/blog/archives/115

上面的文章虽然讲的很详细 不过还是没看太懂

看 了下面的公式和讲解 思想明了

简单KMP练习http://poj.org/problem?id=3461

View Code

 1 #include<stdio.h>

 2 #include<string.h>

 3 char w[10001],p[1000001];

 4 int next[10001];

 5 long num;

 6 void fnext(char *c1)

 7 {

 8     int i,j,k1;

 9     k1 = strlen(c1);

10     j = -1;

11     next[0] = -1;

12     for(i = 1 ; i < k1 ; i++)

13     {        

14         while(j>-1&&c1[i]!=c1[j+1])

15             j = next[j];

16         if(c1[i]==c1[j+1])

17             j++;

18         next[i] = j;

19     }

20 }

21 void kmp(char *c1,char *c2)

22 {

23     int k1,k2,i,j = -1;

24     k1 = strlen(c1);

25     k2 = strlen(c2);

26     next[0] = -1;

27     for(i = 0 ; i < k2 ; i++)

28     {

29         while(j>-1&&c2[i]!=c1[j+1])

30         j = next[j];

31         if(c2[i]==c1[j+1])

32             j++;

33         if(j==k1-1)

34         {

35             j = next[j];

36             num++;

37         }

38     }

39 }

40 int main()

41 {

42     int i,j,t,k;

43     scanf("%d%*c", &t);

44     while(t--)

45     {

46         num = 0;

47         gets(w);

48         gets(p);

49         fnext(w);

50         kmp(w,p);

51         printf("%ld\n",num);

52     }

53     return 0;

54 }

 http://poj.org/problem?id=3080

KMP+暴力

感觉写复杂了。

View Code

  1 #include<stdio.h>

  2 #include<string.h>

  3 char w[61];

  4 int next[10001];

  5 long num[101];

  6 typedef struct node

  7 {

  8     char c[61];

  9     int k;

 10 }st;

 11 st m[1001];

 12 char p[11][61];

 13 void fnext(char *c1)

 14 {

 15     int i,j,k1;

 16     k1 = strlen(c1);

 17     j = -1;

 18     next[0] = -1;

 19     for(i = 1 ; i < k1 ; i++)

 20     {        

 21         while(j>-1&&c1[i]!=c1[j+1])

 22             j = next[j];

 23         if(c1[i]==c1[j+1])

 24             j++;

 25         next[i] = j;

 26     }

 27 }

 28 int kmp(char *c1,char *c2)

 29 {

 30     int k1,k2,i,j = -1,flag = 0;

 31     k1 = strlen(c1);

 32     k2 = strlen(c2);

 33     next[0] = -1;

 34     for(i = 0 ; i < k2 ; i++)

 35     {

 36         while(j>-1&&c2[i]!=c1[j+1])

 37         j = next[j];

 38         if(c2[i]==c1[j+1])

 39             j++;

 40         if(j==k1-1)

 41         {

 42             flag = 1;

 43             break;

 44         }

 45     }

 46     return flag;

 47 }

 48 int pa(char *x,int n)

 49 {

 50     int y,f = 1;

 51     fnext(x);

 52     for(y = 1; y < n ; y++)

 53         if(!kmp(x,p[y]))

 54         {

 55             f = 0;

 56             break;

 57         }

 58     return f;

 59 }

 60 int main()

 61 {

 62     int i,j,t,k,o,max,n,y,f,pf;

 63     char x[61],str[61];

 64     scanf("%d%*c", &t);

 65     while(t--)

 66     {

 67         o = 0;

 68         int g = 0;

 69         for(i = 0 ; i <= 60 ; i++)

 70             m[i].k = 0;

 71         scanf("%d%*c",&n);

 72         gets(w);

 73         for(i = 1 ; i < n ; i++)        

 74         gets(p[i]);        

 75         for(i = 0 ;i <strlen(w) ; i++)

 76         {

 77             g = 0;

 78             x[g] = w[i];

 79             x[g+1]='\0';        

 80             f = 1;

 81             pf = 0;

 82             if(pa(x,n))

 83             {

 84                 j = i;

 85                 while(pa(x,n)&&j<strlen(w))

 86                 {

 87                     pf = 1;

 88                     j++;

 89                     g++;

 90                     x[g] = w[j];

 91                     x[g+1] = '\0';                    

 92                 }

 93                 if(pf)

 94                     x[g]='\0';

 95                 o++;

 96                 strcpy(m[o].c,x);

 97                 m[o].k = strlen(x);

 98             }

 99         }

100         max = 0;    

101         for(i = 1 ; i <= o ; i++)

102             if(m[i].k>max)

103                 max = m[i].k;

104         for(i = 1; i <= o ; i++)

105             if(m[i].k == max)

106             {

107                 strcpy(str,m[i].c);

108                 break;

109             }

110         for(i = 1; i <= o ; i++)

111             if(m[i].k == max)

112             {

113                 if(strcmp(str,m[i].c)>0)

114                 strcpy(str,m[i].c);

115             }

116         if(max<3)

117             printf("no significant commonalities\n");

118         else

119             puts(str);

120     }

121     return 0;

122 }

补一篇写next函数的 写的很好

http://blog.csdn.net/justlovetao/article/details/6737174