NYOJ 5 Binary String Matching （KMP）

题目5

题目信息

运行结果

本题排行

讨论区

Binary String Matching

时间限制：3000 ms  |  内存限制：65535 KB
难度：3

描述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.
样例输入

3
11
1001110110
101
110010010010001
1010
110100010101011

样例输出

3
0
3

这个题的数据很水，可以使用暴力AC，但是还是使用KMP比较好

1.

//暴力
#include <iostream>
#include <cstring>
using namespace std;
int n;
int ans;
char a[20], b[1010];
int main()
{
cin >> n;
while (n--){
ans = 0;
cin >> a >> b;
int lena = strlen(a);
int lenb = strlen(b);
for (int i = 0; i <= lenb - lena; i++){
for (int j = 0; j < lena; j++){
if (a[j] != b[i + j]){
break;
}
if (j == lena - 1)
ans++;
}
}
cout << ans << endl;
}
return 0;
}

2.

//KMP模板
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 1010;
int n;
int next[maxn];
char s[maxn], t[maxn];
int slen, tlen;

void getNext()
{
int j = 0, k = -1;
next[0] = -1;
while (j < slen){
if (k == -1 || s[j] == s[k])
next[++j] = ++k;
else
k = next[k];
}
}

int kmp_count()
{
int ans = 0;
int i, j = 0;
if (slen == 1 && tlen == 1){
if (s[0] == t[0])
return 1;
else
return 0;
}
getNext();
for (i = 0; i < tlen; i++){
while (j > 0 && t[i] != s[j])
j = next[j];
if (s[j] == t[i])
j++;
if (j == slen){
ans++;
j = next[j];
}
}
return ans;
}

int main()
{
cin >> n;
while (n--){
cin >> s >> t;
slen = strlen(s);
tlen = strlen(t);
cout << kmp_count() << endl;
}
return 0;
}