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
【分析】
模板
1 /*
2 宋代李之仪
3 卜算子·我住长江头
4 我住长江头，君住长江尾。日日思君不见君，共饮长江水。
5 此水几时休，此恨何时已。只愿君心似我心，定不负相思意。
6 */
7 #include <iostream>
8 #include <cstdio>
9 #include <algorithm>
10 #include <cstring>
11 #include <vector>
12 #include <utility>
13 #include <iomanip>
14 #include <string>
15 #include <cmath>
16 #include <queue>
17 #include <assert.h>
18 #include <map>
19 #include <ctime>
20 #include <cstdlib>
21 #include <stack>
22 #define LOCAL
23 const int MAXN = 1000000 + 10;
24 const int INF = 100000000;
25 const int SIZE = 450;
26 const int MAXM = 1000000 + 10;
27 const int maxnode = 0x7fffffff + 10;
28 using namespace std;
29 int l1, l2;
30 char a[MAXN], b[MAXN];
31 int next[20000];//不用开太大了..
32 void getNext(){
33 //初始化next数组
34 next[1] = 0;
35 int j = 0;
36 for (int i = 2; i <= l1; i++){
37 while (j > 0 && a[j + 1] != a[i]) j = next[j];
38 if (a[j + 1] == a[i]) j++;
39 next[i] = j;
40 }
41 return;
42 }
43 int kmp(){
44 int j = 0, cnt = 0;
45 for (int i = 1; i <= l2; i++){
46 while (j > 0 && a[j + 1] != b[i]) j = next[j];
47 if (a[j + 1] == b[i]) j++;
48 if (j == l1){
49 cnt++;
50 j = next[j];//回到上一个匹配点
51 }
52 }
53 return cnt;
54 }
55
56 void init(){
57 scanf("%s", a + 1);
58 scanf("%s", b + 1);
59 l1 = strlen(a + 1);
60 l2 = strlen(b + 1);
61 }
62
63 int main(){
64 int T;
65
66 scanf("%d", &T);
67 while (T--){
68 init();
69 getNext();
70 printf("%d\n", kmp());
71 }
72 /*scanf("%s", a + 1);
73 l1 = strlen(a + 1);
74 getNext();
75 for (int i = 1; i <= l1; i++) printf("%d ", next[i]);*/
76 return 0;
77 }