• KMP——模板 练手题


    1. 洛谷  P3375 【模板】KMP字符串匹配

    题目描述

    如题,给出两个字符串s1和s2,其中s2为s1的子串,求出s2在s1中所有出现的位置。

    为了减少骗分的情况,接下来还要输出子串的前缀数组next。如果你不知道这是什么意思也不要问,去百度搜[kmp算法]学习一下就知道了。

    输入输出格式

    输入格式:

    第一行为一个字符串,即为s1(仅包含大写字母)

    第二行为一个字符串,即为s2(仅包含大写字母)

    输出格式:

    若干行,每行包含一个整数,表示s2在s1中出现的位置

    接下来1行,包括length(s2)个整数,表示前缀数组next[i]的值。

    输入输出样例

    输入样例#1:
    ABABABC
    ABA
    输出样例#1:
    1
    3
    0 0 1 

    说明

    时空限制:1000ms,128M

    数据规模:

    设s1长度为N,s2长度为M

    对于30%的数据:N<=15,M<=5

    对于70%的数据:N<=10000,M<=100

    对于100%的数据:N<=1000000,M<=1000

    样例说明:

    所以两个匹配位置为1和3,输出1、3

    血的教训啊,调了一上午的代码,居然错在了输入上,谨记大佬教诲,能不用gets就不用gets!!!

    (⊙v⊙)嗯~ 代码1:

     1 #include<iostream>
     2 #include<cstdio>
     3 #include<cstring>
     4 using namespace std;
     5 
     6 const int N = 1000001;
     7 int n,m,nxt[N],j;
     8 char s1[N],s2[N];
     9 
    10 void INT() {
    11     nxt[0]=0,nxt[1]=0;
    12     for(int i=1; i<m; i++) { 
    13         j = nxt[i];
    14         while(j>0 && s2[i]!=s2[j]) j=nxt[j];
    15         if(s2[i]==s2[j]) nxt[i+1]=j+1;
    16         else nxt[i+1] = 0;
    17     } 
    18 }
    19 
    20 void find() {
    21     j=0;
    22     for(int i=0; i<n; i++) {
    23         while(j>0&&s1[i]!=s2[j])  j=nxt[j];
    24         if(s1[i]==s2[j]) j++;
    25         if(j == m) 
    26             cout<<i-m+2<<endl;
    27     }
    28 }
    29 
    30 int main() {
    31     /*gets(s1);
    32      gets(s2);*/ //无情的删去
    33      cin>>s1>>s2;
    34     n=strlen(s1); m=strlen(s2);
    35     INT();
    36     find();
    37     for(int i=1; i<=m; i++)
    38         cout<<nxt[i]<<" ";
    39     return 0;
    40 }

    (⊙v⊙)嗯~ 代码2:

    算法详解直播间ing

    #include<iostream>
    #include<cstdio>
    #include<cstring> 
    using namespace std;
    
    string s,p;
    int nxt[1000003];
    
    void getnxt(string p) {
        int k=-1,j=0;
        nxt[0]=-1;
        int lenp=p.length();
        while(j<lenp) {
            /**/if(k==-1||p[j]==p[k]) {
                j++;
                //与注释部分作用相同 
                k++;
                nxt[j]=k;
                /*
                 nxt[j]=k+1;
                 k++;
                */
            }
            /**/
            else k=nxt[k];
        }
    }
    
    void kmp(string s,string p) {
        int i=0,j=0;
        int lens=s.length(),
            lenp=p.length();
        while(i<lens) {
            /**/
            if(j==-1||s[i]==p[j]) {
                i++;
                j++;
            }
            /**/
            else
                j=nxt[j];
            if(j==lenp) {
                cout<<i-lenp+1<<endl;
                j=0,i--;
            }
        }
        for(int i=1; i<=lenp; i++) {
            cout<<nxt[i]<<" ";
        }
        return ;
    }
    
    int main() {
        cin>>s;
        cin>>p;
        getnxt(p);
        kmp(s,p);
        return 0;
    }

    不懂的,看这里:

    2. POJ 3461 Oulipo

    Oulipo
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 40182   Accepted: 16143

    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

     
    代码几乎和上一个题一样,因为此题是在②号子串里找①号子串,所以输入的时候倒一下顺序不要用gets输入
     

    (⊙v⊙)嗯~ 代码:

     1 #include<cstdio>
     2 #include<cstring>
     3 #include<iostream>
     4 using namespace std;
     5 char t[1000001],p[1000001];
     6 int lt,lp,f[1000001],ans,s;
     7 void INT()
     8 {
     9     f[1]=0;
    10     for(int i=1;i<lp;i++)
    11     {
    12         int j=f[i];
    13         while(j&&p[i]!=p[j]) j=f[j];
    14         f[i+1]= p[i]==p[j] ? j+1 : 0;
    15     }
    16 }
    17 void Find()
    18 {
    19     int j=0;
    20     for(int i=0;i<lt;i++)
    21     {
    22         while(j&&(p[j]!=t[i])) j=f[j];
    23         if(p[j]==t[i]) j++;
    24         if(j==lp) ans++;
    25     }
    26 }
    27 int main()
    28 {
    29     cin>>s;
    30     while(s--) {
    31         ans=0;
    32         memset(f,0,sizeof(f));
    33         cin>>p>>t;
    34         lt=strlen(t);
    35         lp=strlen(p);
    36         INT();
    37         Find();
    38         cout<<ans<<endl;
    39     }
    40     return 0;
    41 }

    自己选的路,跪着也要走完!

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  • 原文地址:https://www.cnblogs.com/wsdestdq/p/6821229.html
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