写题解前，先放出来一个大神的关于KMP的博客https://blog.csdn.net/v_july_v/article/details/7041827
说真的，昨天学长说今天讲的KMP会有点难，但是跟着听也能理解。果不其然，我中间玩了会QQ，就不会了= = 花了一下午时间去理解，最终明白了三四，但对于非模板题，还是需要一遍又一遍的去更好的理解KMP。
A题 Oulipo
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
题意： 就是给一个字符串，然后给个目标字符串，求这个目标在长的那个字符串里面出现了几次
思路： 本人理解，所谓KMP，就是在暴力匹配的基础上，将每次匹配失败后进行下一匹配的位置进行优化，暴力匹配是每次往右移动一位，但这样会浪费很多时间在匹配过的点上。而KMP中引入了一个next数组，通过目标字符串的相同前缀后缀求出来其next数组，失配时，模式串向右移动的位数为：失配字符所在位置 - 失配字符对应的next 值。
求next数组的模板
有时候也把k初始化为0，j为1
void getnext(int len)
{
int j, k;
j = 0; k = -1; //初始化值为-1且下标从0开始
next[0] = -1;
while(j < len)
{
if(k == -1 || b[j] == b[k])
{
j++; k++;
next[j] = k;
}
else k = next[k]; } }
KMP模板
int kmp(int len1, int len2)
{
int i, j;
int cnt = 0;
i = 0; j = 0;
getnext(len2); //得到目标串的next数组
while(i < len1)
{
if(j == -1 || a[i] == b[j])
{
i++; j++;
}
else j = next[j];
if(j == len2) cnt++; //记录目标串出现的次数
}
return cnt;
}
完整AC代码
#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
const int maxx = 1e6+10;
char s1[maxx],s2[maxx];
int nextt[maxx];
void getnext(int len)
{
int j,k;
j=0;k=-1;
nextt[0] = -1;
while(j<len)
{
if(k == -1 || s2[k] == s2[j]){
++k;++j;
nextt[j] = k;
}
else{k = nextt[k];}
}
}
int kmp(int len1,int len2)
{
int i=0;int j=0;int ans=0;
getnext(len2);
while(i<len1)
{
if(j == -1 || s1[i] == s2[j])
{
++i;++j;
}
else{j = nextt[j];}
if(j==len2){ans++;}
}
return ans;
}
int main()
{
int n;
scanf("%d",&n);
//cin>>n;
while(n--)
{
scanf("%s%s",s2,s1);
//cin>>s2>>s1;
int len1 = strlen(s1);
int len2 = strlen(s2);
printf("%d\n",kmp(len1,len2));
//cout<<kmp(len1,len2)<<endl;
}
return 0;
}
注：
-
cin cout会超时，需要用到scanf和printf，但是c语言的输入输出是不能直接对string类进行操作，方法以后再去学，这里暂时先把string换成了char。
-
求char类型长度用strlen()
求string类型长度用str.length() -
next可能会在vj上报错，改成nextt。