KMP——模板 练手题

Posted 2020-09-17 橘生淮南终洛枳

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 #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 }

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

BAPC 2006 Qualification

 

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！