NYOJ - 5:Binary String Matching

Binary String Matching
出典:NYOJ
ラベル:文字列、文字列マッチング、KMPアルゴリズム
参考資料:
類似タイトル:
タイトル
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
入力
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.
しゅつりょく
For each case, output a single line consist a single integer, tells how many times do B appears as a substring of A.
入力サンプル
3 11 1001110110 101 110010010010001 1010 110100010101011
出力サンプル
3 0 3
テーマの大意
シリアルBにシリアルAをマッチングし、マッチング成功回数を出力する.
問題を解く構想.
KMPアルゴリズムを用いて解く.
リファレンスコード

#include
#include
#define MAXN 1005

char A[MAXN],B[MAXN];
int f[MAXN];
int ans;

void getFail(char *P, int *f){
	int m=strlen(P);
	f[0]=f[1]=0;//      
	for(int i=1;i<m;i++){
		int j=f[i];
		while(j && P[i]!=P[j]) j=f[j];
		f[i+1]= P[i]==P[j]? j+1: 0;
	}
}
 
int find(char *T, char *P, int *f){
	int n=strlen(T),m=strlen(P);
	getFail(P,f);
	int j=0;//      ，   0    
	for(int i=0;i<n;i++){//        
		while(j && P[j]!=T[i]) j=f[j];//      ，      
		if(P[j]==T[i])j++;
		if(j==m) ans++;
	}
}


int main(){
	int N;
	scanf("%d",&N);
	while(N--){
		ans=0;
		scanf("%s%s",A,B);
		find(B,A,f);
		printf("%d
",ans);
	}
	return 0;
}