| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 39112 | Accepted: 15748 |
Description
The French author Georges Perec (1936–1982) once wrote a book, La disparition, without the letter 'e'. He was a member of the Oulipo group. A quote from the book:
Tout avait Pair normal, mais tout s’affirmait faux. Tout avait Fair normal, d’abord, puis surgissait l’inhumain, l’affolant. Il aurait voulu savoir où s’articulait l’association qui l’unissait au roman : stir son tapis, assaillant à tout instant son imagination, l’intuition d’un tabou, la vision d’un mal obscur, d’un quoi vacant, d’un non-dit : la vision, l’avision d’un oubli commandant tout, où s’abolissait la raison : tout avait l’air normal mais…
Perec would probably have scored high (or rather, low) in the following contest. People are asked to write a perhaps even meaningful text on some subject with as few occurrences of a given “word” as possible. Our task is to provide the jury with a program that counts these occurrences, in order to obtain a ranking of the competitors. These competitors often write very long texts with nonsense meaning; a sequence of 500,000 consecutive 'T's is not unusual. And they never use spaces.
So we want to quickly find out how often a word, i.e., a given string, occurs in a text. More formally: given the alphabet {'A', 'B', 'C', …, 'Z'} and two finite strings over that alphabet, a word W and a text T, count the number of occurrences of W in T. All the consecutive characters of W must exactly match consecutive characters of T. Occurrences may overlap.
Input
The first line of the input file contains a single number: the number of test cases to follow. Each test case has the following format:
- One line with the word W, a string over {'A', 'B', 'C', …, 'Z'}, with 1 ≤ |W| ≤ 10,000 (here |W| denotes the length of the string W).
- One line with the text T, a string over {'A', 'B', 'C', …, 'Z'}, with |W| ≤ |T| ≤ 1,000,000.
Output
For every test case in the input file, the output should contain a single number, on a single line: the number of occurrences of the word W in the text T.
Sample Input
3 BAPC BAPC AZA AZAZAZA VERDI AVERDXIVYERDIAN
Sample Output
1 3 0
Source
题意：求一个串在另一个串中出现的次数。思路：
①MP算法，区别于BF算法那样的暴力循环，当遇到不匹配的字符时，不需要回到模式串的开头重新匹配，只需从next处继续即可，如下图：遇到a和b不等，但黑色部分A和B是一样的话，直接从A的右边处继续匹配，next数组需要预处理。
# include <stdio.h>
# include <string.h>
char s[10001], t[1000001];
int next[10001];//失配指针
int slen, tlen;
void init()
{
memset(next, 0, sizeof(next));
next[0] = next[1] = 0;
for(int i=1; i<slen; ++i)
{
int j = next[i];
while(j && s[i] != s[j]) j = next[j];
next[i+1] = s[i]==s[j]?j+1:0;
}
}
int kmp()
{
int ans = 0, j = 0;
for(int i=0; i<tlen; ++i)
{
while(j && t[i] != s[j]) j = next[j];
if(t[i] == s[j]) ++j;
if(j == slen)
{
++ans;//计数器
j = next[j];
}
}
return ans;
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
scanf("%s%s",s,t);
slen = strlen(s);
tlen = strlen(t);
init();
printf("%d\n",kmp());
}
return 0;
}
②KMP实现：区别于MP，多开一个数组用于优化，失配指针更精确。
# include <stdio.h>
# include <string.h>
char s[10001], t[1000001];
int next[10001];//优化后的失配指针
int next2[10001];//原来的失配指针
int slen, tlen;
void init()
{
memset(next, 0, sizeof(next));
memset(next2, 0, sizeof(next2));
next[0] = next[1] = 0;
next2[0] = next2[1] = 0;
for(int i=1; i<slen; ++i)
{
int j = next2[i];
while(j && s[i] != s[j]) j = next[j];
next2[i+1] = next[i+1] = s[i]==s[j]?j+1:0;
if(next[i+1]==j+1 && s[i+1]==s[j+1])//这种情况下，指针可以继续往前跳。
next[i+1] = next[j+1];
}
}
int kmp()
{
int ans = 0, j = 0;
for(int i=0; i<tlen; ++i)
{
while(j && t[i] != s[j]) j = next[j];
if(t[i] == s[j]) ++j;
if(j == slen)
{
++ans;//计数器
j = next[j];
}
}
return ans;
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
scanf("%s%s",s,t);
slen = strlen(s);
tlen = strlen(t);
init();
printf("%d\n",kmp());
}
return 0;
}
