<九度 OJ>题目1094：String Matching

题目描述：

    Finding all occurrences of a pattern in a text is a problem that arises frequently in text-editing programs. 

    Typically,the text is a document being edited,and the pattern searched for is a particular word supplied by the user.  

    We assume that the text is an array T[1..n] of length n and that the pattern is an array P[1..m] of length m<=n.We further assume that the elements of P and  T are all alphabets(∑={a,b...,z}).The character arrays P and T are often called strings of characters.  

    We say that pattern P occurs with shift s in the text T if 0<=s<=n and T[s+1..s+m] = P[1..m](that is if T[s+j]=P[j],for 1<=j<=m).  

    If P occurs with shift s in T,then we call s a valid shift;otherwise,we calls a invalid shift. 

    Your task is to calculate the number of vald shifts for the given text T and p attern P.
输入：

   For each case, there are two strings T and P on a line,separated by a single space.You may assume both the length of T and P will not exceed 10^6. 

输出：

    You should output a number on a separate line,which indicates the number of valid shifts for the given text T and pattern P.
样例输入：

abababab abab

样例输出：

3

KMP算法的简单应用

<span style="font-size:12px;">#include <iomanip>//小数点精确
#include <cstdio>
#include <cstdlib>
#include "vector"
#include "string"
#include "algorithm"
#include <iostream>
#include "stack"
#include "math.h"

using namespace std;

//计算模式P的部分匹配值，保存在next数组中
void MakeNext(const string &P, vector<int> &next)
{
int q, k;//k记录所有前缀的对称值
int m = P.size();//模式字符串的长度
next[0] = 0;//首字符的对称值肯定为0
for (q = 1, k = 0; q < m; ++q)//计算每一个位置的对称值
{
//k总是用来记录上一个前缀的最大对称值
while (k > 0 && P[q] != P[k])
k = next[k - 1];//k将循环递减，值得注意的是next[k]<k总是成立
if (P[q] == P[k])
k++;//增加k的唯一方法
next[q] = k;//获取最终值
}
}

int KmpMatch(const string &T, const string &P, vector<int> &next)
{
int n = 0, m = 0, count = 0;
n = T.size();
m = P.size();
MakeNext(P, next);
for (int i = 0, q = 0; i < n; ++i)
{
while (q > 0 && P[q] != T[i])
q = next[q - 1];
if (P[q] == T[i])
q++;
if (q == m)
{
q = next[q - 1];//寻找下一个匹配
count++;
}
}
return count;
}

int main()
{
string T;//文本
string P;//待搜索字符串
vector<int> next(200, 0);//保存待搜索字符串的部分匹配表（所有前缀函数的对称值）
while (cin >> T >> P)
{//本题为kmp算法的简单应用
cout << KmpMatch(T, P, next) << endl;
}

return 0;
}
/**************************************************************
Problem: 1094
User: EbowTang
Language: C++
Result: Accepted
Time:160 ms
Memory:3052 kb
****************************************************************/</span>

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原文地址：http://blog.csdn.net/ebowtang/article/details/43567601

原作者博客：http://blog.csdn.net/ebowtang