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
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！