Oulipo
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
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
Sample Output
Source
题意：找出第一个串在第二个串出现次数；
思路：kmp板子；
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
#include<stdlib.h>
#include<time.h>
#include<bits/stdc++.h>
using namespace std;
#define LL long long
#define pi (4*atan(1.0))
#define eps 1e-6
#define bug(x) cout<<"bug"<<x<<endl;
const int N=5e4+10,M=1e6+10,inf=1e9+10;
const LL INF=5e17+10,mod=1e9+7;
void makeNext(const char P[],int next[],int strp)
{
int q,k;
//int m = strlen(P);
next[0] = 0;
for (q = 1,k = 0; q < strp; ++q)
{
while(k > 0 && P[q] != P[k])
k = next[k-1];
if (P[q] == P[k])
{
k++;
}
next[q] = k;
}
}
int kmp(const char T[],const char P[],int next[],int strt,int strp)
{
int n,m;
int i,q;
int ans=0;
//n = strlen(T);
//m = strlen(P);
makeNext(P,next,strp);
for (i = 0,q = 0; i < strt; ++i)
{
while(q > 0 && P[q] != T[i])
q = next[q-1];
if (P[q] == T[i])
{
q++;
}
if (q == strp)
{
ans++;
q=next[q-1];
}
}
return ans;
}
char T[1000010];
char P[10010];
int net[10010];
int main()
{
int TT;
scanf("%d",&TT);
while(TT--)
{
memset(net,0,sizeof(net));
scanf("%s%s",P,T);
int strp=strlen(P);
int strt=strlen(T);
int ans=kmp(T,P,net,strt,strp);
printf("%d\n",ans);
}
return 0;
}
hdu 1686 Oulipo kmp
原文：http://www.cnblogs.com/jhz033/p/6939713.html
