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poj3461 Oulipo
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
昨天去做cf的时候B题写KMP的时候调了半天调不出来真是打脸……
所以现在开始疯狂刷KMP(当复习吧)
这题当模板用
#include<cstdio>#include<iostream>#include<cstring>#include<cstdlib>#include<algorithm>#include<cmath>#include<queue>#include<deque>#include<set>#include<map>#include<ctime>#define LL long long#define inf 0x7ffffff#define pa pair<int,int>#define pi 3.1415926535897932384626433832795028841971using namespace std;inline LL read(){ LL x=0,f=1;char ch=getchar(); while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();} while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();} return x*f;}inline void write(LL a){ if (a<0){printf("-");a=-a;} if (a>=10)write(a/10); putchar(a%10+‘0‘);}inline void writeln(LL a){write(a);printf("\n");}int next[100010];char s[10010],p[1000010];int ls,lp;inline void pre(){ memset(next,0,sizeof(next)); int j=0; for (int i=2;i<=ls;i++) { while (j>0 && s[j+1]!=s[i])j=next[j]; if (s[j+1]==s[i])j++; next[i]=j; }}inline void KMP(){ int j=0,ans=0; for (int i=1;i<=lp;i++) { while (j>0 && s[j+1]!=p[i])j=next[j]; if (s[j+1]==p[i])j++; if (j==ls) { ans++; j=next[j]; } } printf("%d\n",ans);}inline void work(){ scanf("%s",s+1); scanf("%s",p+1); ls=strlen(s+1);lp=strlen(p+1); pre(); KMP();}int main(){ int T=read(); while (T--)work();}
poj3461 Oulipo