• hdu 1686 Oulipo kmp


    Oulipo

    Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)



    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
    3 BAPC BAPC AZA AZAZAZA VERDI AVERDXIVYERDIAN
     
    Sample Output
    1 3 0
     
    Source
     

    题意:找出第一个串在第二个串出现次数;

    思路:kmp板子;

    #pragma comment(linker, "/STACK:1024000000,1024000000")
    #include<iostream>
    #include<cstdio>
    #include<cmath>
    #include<string>
    #include<queue>
    #include<algorithm>
    #include<stack>
    #include<cstring>
    #include<vector>
    #include<list>
    #include<set>
    #include<map>
    #include<stdlib.h>
    #include<time.h>
    #include<bits/stdc++.h>
    using namespace std;
    #define LL long long
    #define pi (4*atan(1.0))
    #define eps 1e-6
    #define bug(x)  cout<<"bug"<<x<<endl;
    const int N=5e4+10,M=1e6+10,inf=1e9+10;
    const LL INF=5e17+10,mod=1e9+7;
    
    void makeNext(const char P[],int next[],int strp)
    {
        int q,k;
        //int m = strlen(P);
        next[0] = 0;
        for (q = 1,k = 0; q < strp; ++q)
        {
            while(k > 0 && P[q] != P[k])
                k = next[k-1];
            if (P[q] == P[k])
            {
                k++;
            }
            next[q] = k;
        }
    }
    
    int kmp(const char T[],const char P[],int next[],int strt,int strp)
    {
        int n,m;
        int i,q;
        int ans=0;
        //n = strlen(T);
        //m = strlen(P);
        makeNext(P,next,strp);
        for (i = 0,q = 0; i < strt; ++i)
        {
            while(q > 0 && P[q] != T[i])
                q = next[q-1];
            if (P[q] == T[i])
            {
                q++;
            }
            if (q == strp)
            {
                ans++;
                q=next[q-1];
            }
        }
        return ans;
    }
    char T[1000010];
    char P[10010];
    int net[10010];
    int main()
    {
        int TT;
        scanf("%d",&TT);
        while(TT--)
        {
            memset(net,0,sizeof(net));
            scanf("%s%s",P,T);
            int strp=strlen(P);
            int strt=strlen(T);
            int ans=kmp(T,P,net,strt,strp);
            printf("%d
    ",ans);
        }
    
    
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/jhz033/p/6939713.html
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