标签:
abcde a3 aaaaaa aa #
0 3
题意:最多能从文本串中剪出多少个模式串。
因为不能重叠,所以当j匹配的模式串的末尾并且匹配成功时直接把j变成0从新匹配就
好了。
#include <cstring>
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <vector>
#include <cmath>
using namespace std;
#define maxn 1111
char P[maxn], T[maxn];
int n, m;
#define next Next
int next[maxn];
void get_next (char *p) {
int t;
t = next[0] = -1;
int j = 0;
while (j+1 < m) {
if (t < 0 || p[j] == p[t]) {//匹配
j++, t++;
next[j] = t;
}
else //失配
t = next[t];
}
}
int kmp () {
int ans = 0;
get_next (P);
int i = 0, j = 0;
while (i < n && j < m) {
if (j < 0 || T[i] == P[j]) {
if (j == m-1) {
ans++;
i++;
j = 0;
continue;
}
i++, j++;
}
else {
j = next[j];
}
}
return ans;
}
int main () {
while (cin >> T) {
if (T[0] == '#' && strlen (T) == 1)
break;
cin >> P;
n = strlen (T);
m = strlen (P);
printf ("%d\n", kmp ());
}
return 0;
}标签:
原文地址:http://blog.csdn.net/morejarphone/article/details/51348602