撰写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.
InputThe 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.
OutputFor 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：找出文本串中可部分重叠的模式串个数
题解：KMP模板
代码：
#include <iostream>
#include <cstring>
#include <string>
#include <vector>
using namespace std;
const int N = 1000002;
int nextt[N];
char S[N],T[N];
//vector<int>S,T;
int slen, tlen;
void getNext()
{
int j, k;
j = 0; k = -1; nextt[0] = -1;
while(j < tlen)
if(k == -1 || T[j] == T[k])
nextt[++j] = ++k;
else
k = nextt[k];
}
/*
返回模式串T在主串S中首次出现的位置
返回的位置是从0开始的。
*/
int KMP_Index()
{
int i = 0, j = 0,ans=0;//ans=-1;
getNext();
while(i < slen && j < tlen)
{
if(j == -1 || S[i] == T[j])
{
i++; j++;
}
else
j = nextt[j];
if(j == tlen){
ans++;//ans=i-j;
//j=0;//break;
j=nextt[j];
}
}
return ans;
}
int main()
{
int TT;
int i, cc;
cin>>TT;
while(TT--)
{
scanf("%s%s", T,S);
slen = strlen(S);
tlen = strlen(T);
cout<<KMP_Index()<<endl;
}
return 0;
}