Oulipo (HDU - 1686)

题面

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.

输入

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.

输出

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.

样例输入

13
2BAPC
3BAPC
4AZA
5AZAZAZA
6VERDI
7AVERDXIVYERDIAN

样例输出

11
23
30

提示

无

思路

KMP模板题

代码

 1char s[mxn], t[mxn];
 2int nxt[mxn];
 3
 4void getnxt(char* t, int m)
 5{
 6    int i = 0, j = -1; nxt[0] = -1;
 7    while (i < m)
 8    {
 9        if (j == -1 || t[i] == t[j]) {
10            i++, j++;
11            // if (t[i] == t[j])
12            //     nxt[i] = nxt[j]; // next数组优化
13            // else
14                nxt[i] = j;
15        } else
16            j = nxt[j];
17    }
18}
19
20int KMP(char* s, char* t, int n, int m)
21{
22    int i = 0, j = 0, ans = 0;
23    while (i < n)
24    {
25        if (j == -1 || s[i] == t[j]) {
26            i++, j++;
27            if (j >= m) {   // 匹配
28                ans++;
29                j = nxt[j];
30                // return i-j;
31            }
32        } else
33            j = nxt[j];
34    }
35    return ans;
36    // return -1;
37}
38
39int main()
40{
41    int T; scanf("%d", &T);
42    while(T--)
43    {
44        scanf("%s%s", t, s);
45        int tl = strlen(t), sl = strlen(s);
46        getnxt(t, tl);
47        printf("%d\n", KMP(s, t, sl, tl));
48    }
49    return 0;
50}