http://acm.hdu.edu.cn/showproblem.php?pid=1686
Oulipo
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6098 Accepted Submission(s): 2448
Problem 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
1 #include <iostream>
2 #include <stdlib.h>
3 #include <stdio.h>
4 #include <cstring>
5 using namespace std;
6 int n,m,nxt[10005],kk,t;
7 char b[10005],a[1000005];
8 ///此题在基础的kmp上加了多次匹配。
9 ///就意味着我们在匹配完一次字串后，要跳到最适合的位置,继续查找
10 ///继续利用kmp的思想。某些位置已经匹配过，就不要匹配了。
11 /// xxxxxxxabbaab*xxxxxx
12 /// abbaaba
13 //我们跳跃之后的位置 abbaaba 而跳跃的位置与next数组有关
14 void buildnxt()
15 {
16 int j,k;
17 m=strlen(b);
18 nxt[0]=-1;
19 j=0;k=-1;
20 while(j<m)
21 {
22 if((k==-1)||b[j]==b[k])
23 {
24 j++;
25 k++;
26 nxt[j]=k;
27 }
28 else k=nxt[k];
29 }
30 }
31 int kmp()
32 {
33 int k=0,l=0,cou=0;
34 n=strlen(a);
35 int ans=m,kk=nxt[m];///ans在字串中下标，和起点距离ans+1
36 while(1)
37 {
38 if(kk!=0&&kk!=-1) {ans=kk;kk=nxt[ans];}
39 else break;
40 }///要找最小的跳跃点,所以从next尾端返回去找到首个非负值。
41 while(k<n)
42 {
43 if((l==-1)||a[k]==b[l])
44 {
45 k++;
46 l++;
47 }
48 else l=nxt[l];
49 if(l==m)
50 {
51 cou++;
52 if(kk==0) continue;///如果是尾端next数组是0的话，主串中匹配的子串中没有重复。
53 ///也就是说在匹配的主串中，没有可以跳跃的点。
54 if(k==n-1) break;///如果k已经是主串末尾了，就不能还有继续可以匹配的字串了。
55 k=k-l+ans;///k-l(起点)+ans
56 l=0;
57 }
58 }
59 return cou;
60 }
61 int main()
62 {
63 scanf("%d",&t);
64 getchar();
65 while(t--)
66 {
67 gets(b);
68 gets(a);
69 memset(nxt,0,sizeof(nxt));
70 buildnxt();
71 printf("%d\n",kmp());
72 }
73 return 0;
74 }