POJ3461

Oulipo

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 24338 Accepted: 9763

题目链接：http://poj.org/problem?id=3461

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

解题思路：

题意是求在文本串t里查找有多少个模式串w，看看n和m的范围，BF算法必超时，所以自然想到KMP算法·······

对于KMP算法，不想做太多讲解，因为我感觉有一篇博客写的比我好很多，拜读后我也受益匪浅。而后又翻阅了导论、刘汝佳蜀黍的厚书······终于找到了与之相匹配的代码······

本题不是问是否存在，而是问存在多少个，所以在第一次匹配完成后，模式串要进行回溯，继续下一次匹配。

（附链接：http://kb.cnblogs.com/page/176818/）

完整代码：

#include <functional>#include <algorithm>#include <iostream>#include <fstream>#include <sstream>#include <iomanip>#include <numeric>#include <cstring>#include <climits>#include <cassert>#include <complex>#include <cstdio>#include <string>#include <vector>#include <bitset>#include <queue>#include <stack>#include <cmath>#include <ctime>#include <list>#include <set>#include <map>using namespace std;#pragma comment(linker, "/STACK:102400000,102400000")typedef long long LL;typedef double DB;typedef unsigned uint;typedef unsigned long long uLL;/** Constant List .. **/ //{const int MOD = int(1e9)+7;const int INF = 0x3f3f3f3f;const LL INFF = 0x3f3f3f3f3f3f3f3fLL;const DB EPS = 1e-9;const DB OO = 1e20;const DB PI = acos(-1.0); //M_PI;string w , t;int next[10001];void getNext(){    int lenw = w.length();    next[0] = 0 ;    next[1] = 0;    for(int i = 1 ; i < lenw ; i ++)    {        int j = next[i];        while(j && w[i] != w[j])            j = next[j];        next[i+1] = w[i] == w[j] ? j + 1 : 0;    }}int kmp(){    int lent = t.length();    int lenw = w.length();    int ans = 0;    getNext();    int j = 0;    for(int i = 0 ; i < lent ; i++)    {        while(j && w[j] != t[i])            j = next[j];        if(t[i] == w[j])            j ++;        if(j == lenw)        {            ans++;            j = next[j];        }    }    return ans;}int main(){    #ifdef DoubleQ    freopen("in.txt","r",stdin);    #endif    int T;    scanf("%d",&T);    while(T--)    {        cin >> w >>t;        printf("%d\n",kmp());    }}