poj 3461 Oulipo(KMP 字符串匹配算法)

Oulipo
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Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 22354 Accepted: 8922

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 wordW and a text T, count the number of occurrences of W inT. All the consecutive characters of W must exactly match consecutive characters ofT. 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 stringW).
• 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 wordW in the text T.

Sample Input

3BAPCBAPCAZAAZAZAZAVERDIAVERDXIVYERDIAN

Sample Output

130题目：输入两个字符串，输出，第二个串里有多少个和第一个匹配的子串；题目给的数据量较大（用 KMP算法 ，建议scanf，printf输入输出），用朴素的字符串匹配算法会超时的；KMP算法：（源自百度百科）
基本思想
假设在模式匹配的进程中，执行T[i]和W[j]的匹配检查。若T[i]=W[j]，则继续检查T[i+1]和W[j+1]是否匹配。若T[i]<&gt;W[j]，则分成两种情况：若j=1，则模式串右移一位，检查T[i+1]和W[1]是否匹配；若1<j<=m，则模式串右移j-next(j）位，检查T[i]和W[next(j)]是否匹配。重复此过程直到j=m或i=n结束；主要是next[]数组的求解（KMP主要是通过next数组实现跳跃多个字符，提高效率的）；关于next数组的求解请看转载的博客KMP算法总结代码：
#include <iostream>#include<string.h>#define N 1000001using namespace std;int next[N];char str[N],s[N];void get_next()//获取next数组{    unsigned int i, t,length=strlen(s);    i = 1;    t = 0;    next[1] = 0;    while(i < length + 1)    {        while(t > 0 && s[i - 1] != s[t - 1])        {            t = next[t];        }        ++t;        ++i;        if(s[i - 1] == s[t - 1])        {            next[i] = next[t];        }        else        {            next[i] = t;        }    }    //s末尾的结束符控制，用于寻找目标字符串中的所有匹配结果用    while(t > 0 && s[i - 1] != s[t - 1])    {        t = next[t];    }    ++t;    ++i;    next[i] = t;}int KMP(){    int i,j,n;    int s_length=strlen(s);    int text_length=strlen(str);    get_next();//Next(s);    i = 0;    j = 1;    n = 0;    while(s_length + 1 - j <= text_length - i)    {        if(str[i] == s[j - 1])        {            ++i;            ++j;            if(j==s_length + 1)//匹配则计数加一            {               n++;                j = next[j];            }        }        else//匹配失败则跳到next位置，重新匹配        {            j = next[j];            if(j==0)            {                ++i;                ++j;            }        }    }    return n;}int main(){    int t,k;    while(cin>>t)    {        while(t--)        {            memset(str,'\0',sizeof(str[0])*124);            memset(s,'\0',sizeof(s[0])*124);            cin>>(s);            cin>>(str);            k=KMP();            cout<<k<<endl;        }    }    return 0;}