hdu_1686 Oulipo

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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
3BAPCBAPCAZAAZAZAZAVERDIAVERDXIVYERDIAN
 

Sample Output
130
这题与常规kmp算法稍微改变一点就行了。首先next数组要多保存一个值,存储最后一个字符的匹配情况。
另外当字符穿完全匹配时。不要立即返回,而应该让统计次数+1后,根据next数组最后一个值,继续寻找匹配字符串。
#include<iostream>#include<cstdio>#include<cstring>using namespace std;#define N 10001#define M 1000001char s1[N];char s2[M];void GetNext(const char s[],int next[]){    int k = -1,j = 0;    next[0] = -1;    while(j < (int)strlen(s1)){        if(k == -1 || s[k] == s[j]){            ++k;            ++j;            next[j] = k;        }        else k = next[k];    }}int KMP(const char s1[],const char s2[]){    int l1 = strlen(s1);    int l2 = strlen(s2);    int next[l1+1];    GetNext(s1,next);    int i, j, n = 0;    i = 0;    j = 0;    while(i < l2){        if(j == -1 || s1[j] == s2[i]){            ++i;            ++j;        }        else j = next[j];        if(j == l1){            ++n;            j = next[j];        }    }    return n;}int main(){    int n;    scanf("%d",&n);    while(n--){        scanf("%s%s",s1,s2);        printf("%d\n",KMP(s1,s2));    }}


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