| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 36903 | Accepted: 14898 |
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
/* 题意：给你两个字符串P,T让你判断P在T中出线了多少次 */ /* next数组就是表示以第i位为结尾的字符串，前缀后缀中最长相等字符串的长度 */ #include<stdio.h> #include<iostream> #include<string.h> #define M 1000010 using namespace std; int next[M],sum,cur=0; void makeNext(const char P[],int next[]) { int q,k;//q:模版字符串下标；k:最大前后缀长度 int m = strlen(P);//模版字符串长度 next[0]=0;//模版字符串的第一个字符的最大前后缀长度为0 for (q=1,k=0;q<m;++q)//for循环，从第二个字符开始，依次计算每一个字符对应的next值 { /* 这里的while的作用：如果p[q]和p[k]相匹配了，那就说明前q字母组成的字符串最长的长度就是k了，如果没有匹配上的话，就看看前一个 状态最长前后缀长度next[k-1]能不能和p[q]匹配一直这么找下去。 */ while(k>0&&P[q]!=P[k])//递归的求出P[0]···P[q]的最大的相同的前后缀长度k { k=next[k-1]; //不理解没关系看下面的分析，这个while循环是整段代码的精髓所在，确实不好理解 } if (P[q]==P[k])//如果相等，那么最大相同前后缀长度加1 /*仔细想想如果这最后一个字母和遍历到的这个字母都一样了，那么最长的匹配块肯定就会多一个*/ { k++; } next[q]=k; } } int kmp(const char T[],const char P[],int next[]) { int n,m; int i,q; n=strlen(T); m=strlen(P); makeNext(P,next); for(i=0,q=0;i<n;++i) { while(q>0&&P[q]!=T[i]) q=next[q-1]; /*这里的while就是参照next数组中的q位置的值，看看需要向前移动多少个单位*/ if (P[q]==T[i]) { q++; } if(q==m)//找到一个完整的字符串 { cur++; } } } int main() { freopen("C:\Users\acer\Desktop\in.txt","r",stdin); char p[10005],t[1000005]; int n; scanf("%d",&n); while(n--) { scanf("%s %s ",&p,&t); cur=0; memset(next,0,sizeof next); makeNext(p,next); kmp(t,p,next); printf("%d ",cur); } return 0; }