渣渣ACM日记——3461-Oulipo(POJ)

Oulipo

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 28717 Accepted: 11472

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

3BAPCBAPCAZAAZAZAZAVERDIAVERDXIVYERDIAN

Sample Output

130

//第一次学KMP  超时了好多次  学习学习！

AC代码：

#include <iostream>#include <cstdio>#include <string.h>using namespace std;const int MAXN=1000000+50;int next[MAXN];int tlen,plen;int sum;int main(){void GetNext(char *pat);void kmp(char *text,char *pat);char text[MAXN];char pat[10050];int t;cin>>t;while(t--){sum=0;scanf("%s%s",pat,text);tlen=strlen(text);plen=strlen(pat);GetNext(pat);kmp(text,pat); cout<<sum<<endl;}return 0;}void GetNext(char *pat){int j=0,k=-1;next[0]=-1;while(j<plen){if(k==-1||pat[j]==pat[k])next[++j]=++k;else   k=next[k];}}void kmp(char *text,char *pat){int i=0,j=0;while(i<tlen&&j<plen){if(j==-1||text[i]==pat[j]){i++;j++;}else   j=next[j];if(j==plen){  sum++;  j=next[j];}}}

//网上查了都说sunday比较吊 这个很多人都超时  只能用KMP  或者是我们写不对  

//超时的sunday

#include <iostream>#include <cstdio>#include <string.h>using namespace std;int book[30];int plen,tlen;const int MAXN=1000000+20;int sum;int main(){ void getbook(char *pat);void judge(char *text,char *pat);char text[MAXN],pat[10005];int i,t;cin>>t;while(t--){sum=0;    scanf("%s%s",pat,text); plen=strlen(pat),tlen=strlen(text);memset(book,-1,sizeof(book));getbook(pat);judge(text,pat);cout<<sum<<endl;}return 0;}void getbook(char *pat){ int i; for(i=plen-1;i>=0;i--) if(book[pat[i]-'A']==-1) book[pat[i]-'A']=i;    /*for(i=1;i<128;i++)    cout<<i<<"  "<<book[i]<<endl;*/ }void judge(char *text,char *pat){int i,j,now; //j匹配了几个 for(i=0,j=0;i<tlen;){if(text[i]!=pat[j])   //不匹配 { now=i-j+plen;   //子串后面位置对应的母串的位置 if(now<tlen){   //没越界 if(book[text[now]]==-1)//子串后面位置对应的母串没在子串出现 i=now+1;elsei=now-book[text[now]-'A']; //校准下次开始匹配的母串位置 j=0;  //下次匹配从子串首字符开始     }else  //越界 找不到        return ;}else        //匹配一个  { i++,j++; }if(j==plen){    sum++;j=0;  if(i<tlen){   if(book[text[i]]==-1)i++;elsei=i-book[text[i]-'A'];     }else        return ;    }}}