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
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 ",exd_kmp(aa,bb,n,m)); 61 } 62 return 0; 63 }
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 }