NYoj-Binary String Matching-KMP算法
Binary String Matching
时间限制:3000 ms | 内存限制:65535 KB
难度:3
- 描写叙述
- Given two strings A and B, whose alphabet consist only ‘0’ and ‘1’. Your task is only to tell how many times does A appear as a substring of B?
For example, the text string B is ‘1001110110’ while the pattern string A is ‘11’, you should output 3, because the pattern A appeared at the posit
- 输入
- The first line consist only one integer N, indicates N cases follows. In each case, there are two lines, the first line gives the string A, length (A) <= 10, and the second line gives the string B, length (B) <= 1000. And it is guaranteed that B is always longer than A.
- 输出
- For each case, output a single line consist a single integer, tells how many times do B appears as a substring of A.
- 例子输入
-
3 11 1001110110 101 110010010010001 1010 110100010101011
- 例子输出
-
3 0 3
-
/******KMP算法*********/ #include<cstdio> #include<cstring> #include<iostream> #include<cmath> #include<algorithm> //#define MAXSTRLEN 1000 char s[1010]; char t[11]; int next[1010]; using namespace std; void Next() { int i=0; int j=-1; next[0]=-1; int len_t=strlen(t); while(i<len_t-1) { if(j==-1||t[i]==t[j]) { i++; j++; next[i]=j; } else j=next[j]; } } int KMP() { int i=-1; int j=-1; int len_t=strlen(t); int len_s=strlen(s); int count=0; Next(); while(i<len_s) { if(j==-1||s[i]==t[j]) { i++; j++; } else { j=next[j]; } if(j==len_t) { i=i-j; j=-1; count++; } } return count; } int main() { int T; scanf("%d",&T); while(T--) { scanf("%s %s",t,s); printf("%d\n",KMP()); } return 0; }
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