POJ 3461 Oulipo
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 45665 | Accepted: 18233 |
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
Problem Idea
题目理解:给你一个T串和P串,问你P串在T串中出现了多少次。即求解 模板串在待匹配串中出现的次数
接下来的2*N行先后分别给出了每一组测试用例的模板串,和待匹配串。
Source Code
#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
char T[1000000+100];//待匹配串
char P[10000+100];//模板串
int f[10000+101];//失配指针,f[i]表示状态i失配时应该转移到的新状态
//记住这里f要比P多一位,因为P到m-1即可,但是f还要计算出m的失配指针
int cnt;//统计P串在T串中出现了多少次
void find(char* T,char* P,int* f){//找到所有的匹配点
int n=strlen(T);//待匹配串的长度
int m=strlen(P);//模板串的长度
int j=0;//当前节点编号,初始化为0号节点
for(int i=0;i<n;i++){
while (j && T[i]!=P[j]) j=f[j];//顺着失配边走,直到可以匹配
if(T[i]==P[j]) j++;
if(j==m) cnt++;
//printf("%d\n",i-m+1);//找到了,输出待匹配串中匹配位置的下标
}
}
void getFail(char* P,int* f){
int m=strlen(P);
f[0]=f[1]=0;//递推边界的初值
for(int i=1;i<m;i++){//虽然字符串P是从0到m-1的名单上要求出f【m]的值
int j=f[i];
while(j && P[i]!=P[j]) j=f[j];
f[i+1]=P[i]==P[j]?j+1:0;//如果P[i]==P[j],则递推更新f[i+1]
}
}
int main() {
int k;
scanf("%d",&k);
getchar();//必须要getchar();
while(k--){
cnt=0;
scanf("%s %s",P,T);
//getchar();加不加这一行,都可以AC
getFail(P,f);
find(T,P,f);
printf("%d\n",cnt);
}
return 0;
}