NYOJ-5:Binary String Matching
来源:http://acm.nyist.edu.cn/JudgeOnline/problem.php?pid=5
标签:字符串,字符串匹配,KMP算法
参考资料:
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题目
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.
输入样例
3
11
1001110110
101
110010010010001
1010
110100010101011
输出样例
3
0
3
题目大意
在串B中匹配串A,输出匹配成功次数。
解题思路
利用KMP算法求解。
参考代码
#include<stdio.h>
#include<string.h>
#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;//递推边界初值
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;//当前结点编号,初始为0号结点
for(int i=0;i<n;i++){//文本串当前指针
while(j && P[j]!=T[i]) j=f[j];//顺着失配边走,直到可以匹配
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\n",ans);
}
return 0;
}