poj 3461 （看毛片算法初探）

Oulipo
Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%lld & %llu
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Practice

POJ 3461
Appoint description:
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

KMP模板题了

#include <cstdio>#include <cstring>#define M 10010#define N 1000010char str[N], ptr[M];//str 主串，ptr 模板串int next[M], plen, slen, num;//next[i]就是存储当模式串在第 i 处失配时，下次该匹配的位置//next[i] = -1 的意思是模式串与主串应该重新在 i+1 地方重新匹配void getnext(){    int i = 0, k = -1;    next[0] = -1;    while(i < plen)    {        if(k == -1 || ptr[i] == ptr[k])//匹配成功        {            i++, k++;            if(ptr[i] == ptr[k])            {                next[i] = next[k];            }            else            {                next[i] = k;//由前缀与后缀匹配的情况，得到此处失配时应该再次对比的位置            }        }        else        {            k = next[k];//失配回溯，这个地方很难理解，可以认为回到了一个前缀与后缀可以相等的地方，或者直接回到了 -1 重新匹配的点            //在这个地方想不通呢，对比一下getnext函数与kmp函数，可以发现这个相当于又回到了求主串与模板串的地方，模式串就等于主串，匹配不成功就next            //就理解为把当前的模式串看成主串，回溯到得那个串看成子串，然后再次匹配，一直循环，知道 k == -1；跳过，进入下一段        }    }}void kmp(){    int pi = 0, si = 0;//从开始的地方比较    while(si < slen)    {        if(pi == -1 || str[si] == ptr[pi])//匹配成功        {            pi++, si++;        }        else//匹配失败        {            pi = next[pi];//回溯到应该重新匹配的位置        }        if(pi == plen)//完全匹配成功        {            pi = next[pi];//回溯            num++;//计数        }    }    printf("%d\n", num);}int main(){    int t;    scanf("%d", &t);    while(t--)    {        scanf("%s", ptr);        scanf("%s", str);        slen = strlen(str);        plen = strlen(ptr);        num = 0;        getnext();        kmp();    }    return 0;}