HDUOJ------1711Number Sequence
Number Sequence
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 9114 Accepted Submission(s): 4166
Problem Description
Given two sequences of numbers : a[1], a[2], ...... , a[N], and b[1], b[2], ...... , b[M] (1 <= M <= 10000, 1 <= N <= 1000000). Your task is to find a number K which make a[K] = b[1], a[K + 1] = b[2], ...... , a[K + M - 1] = b[M]. If there are more than one K exist, output the smallest one.
Input
The first line of input is a number T which indicate the number of cases. Each case contains three lines. The first line is two numbers N and M (1 <= M <= 10000, 1 <= N <= 1000000). The second line contains N integers which indicate a[1], a[2], ...... , a[N]. The third line contains M integers which indicate b[1], b[2], ...... , b[M]. All integers are in the range of [-1000000, 1000000].
Output
For each test case, you should output one line which only contain K described above. If no such K exists, output -1 instead.
Sample Input
2
13 5
1 2 1 2 3 1 2 3 1 3 2 1 2
1 2 3 1 3
13 5
1 2 1 2 3 1 2 3 1 3 2 1 2
1 2 3 2 1
Sample Output
6
-1
Source
HDU 2007-Spring Programming Contest
kmp 基础
代码：
1 /*@kmp扩展@龚细军*/
2 #include<stdio.h>
3 #include<string.h>
4 int aa[1000004],bb[10005];
5 int next[10005];
6 //依旧使用next数组
7 void get_next(int *pt,int len)
8 {
9 memset(next,0,sizeof(next));
10 int i=0,j=-1;
11 next[0]=-1;
12 while(i<len)
13 {
14 if(j==-1||pt[i]==pt[j])
15 {
16 ++i;
17 ++j;
18 if(pt[i]!=pt[j])
19 next[i]=j;
20 else
21 next[i]=next[j];
22 }
23 else
24 j=next[j];
25 }
26 }
27 //kmp的扩展
28 int exd_kmp(int *ps,int *pt,int lens,int lent)
29 {
30 int i=-1,j=-1;
31 get_next(pt,lent);
32 while(i<lens)
33 {
34 if(j==-1||ps[i]==pt[j])
35 {
36 ++i;
37 ++j;
38 }
39 else
40 j=next[j];
41 if(j==lent)break;
42 }
43 if(j==lent)
44 return i-j+1;
45 else
46 return -1;
47 }
48
49 int main()
50 {
51 int test,n,m,i;
52 scanf("%d",&test);
53 while(test--)
54 {
55 scanf("%d%d",&n,&m);
56 for(i=0;i<n;i++)
57 scanf("%d",&aa[i]);
58 for(i=0;i<m;i++)
59 scanf("%d",&bb[i]);
60 printf("%d\n",exd_kmp(aa,bb,n,m));
61 }
62 return 0;
63 }java代码：
1 import java.util.Scanner;
2
3
4 public class Main {
5
6 public static void main(String args [])
7 {
8 mt aa = new mt();
9 Scanner reader =new Scanner(System.in);
10 int test=reader.nextInt();
11 while((test--)>0)
12 {
13 aa.lena=reader.nextInt();
14 aa.lenb=reader.nextInt();
15 aa.init(aa.lena+1, aa.lenb+1);
16 for(int i=0;i<aa.lena;i++)
17 aa.a[i]=reader.nextInt();
18 for(int i=0;i<aa.lenb;i++)
19 aa.b[i]=reader.nextInt();
20 kmp(aa);
21 }
22 }
23 private static void kmp(mt aa)
24 {
25 int i,j;
26 aa.next[i=0]=-1;
27 j=-1;
28 while(i<aa.lenb)
29 {
30 if(j==-1||aa.b[i]==aa.b[j])
31 {
32 i++;
33 j++;
34 if(aa.b[i]==aa.b[j])
35 aa.next[i]=aa.next[j];
36 else
37 aa.next[i]=j;
38 }
39 else
40 j=aa.next[j];
41 }
42 i=j=0;
43 while(i<aa.lena&&j<aa.lenb)
44 {
45 if(j==-1||aa.a[i]==aa.b[j])
46 {
47 i++;
48 j++;
49 }
50 else j=aa.next[j];
51 }
52 if(j==aa.lenb)
53 out(i-aa.lenb+1);
54 else
55 out(-1);
56 }
57 private static void out(int aa)
58 {
59 System.out.println(aa);
60 }
61 }
62 class mt
63 {
64 int [] b,a,next;
65 int lena,lenb;
66 void init(int lena,int lenb)
67 {
68 a=new int [lena];
69 b=new int [lenb];
70 next =new int [lenb];
71 }
72 }