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：

 /*@kmp扩展@龚细军*/
 #include<stdio.h>
 #include<string.h>
 int aa[],bb[];
 int next[];
 //依旧使用next数组
 void get_next(int *pt,int len)
 {
     memset(next,,sizeof(next));
     int i=,j=-;
     next[]=-;
     while(i<len)
     {
         if(j==-||pt[i]==pt[j])
         {
          ++i;
          ++j;
          if(pt[i]!=pt[j])
              next[i]=j;
          else
              next[i]=next[j];
         }
         else
             j=next[j];
     }
 }
 //kmp的扩展
 int exd_kmp(int *ps,int *pt,int lens,int lent)
 {
     int i=-,j=-;
     get_next(pt,lent);
     while(i<lens)
     {
         if(j==-||ps[i]==pt[j])
         {
             ++i;
             ++j;
         }
         else
             j=next[j];
      if(j==lent)break;
     }
     if(j==lent)
         return i-j+;
     else
         return -;
 }

 int main()
 {
     int test,n,m,i;
     scanf("%d",&test);
     while(test--)
     {
         scanf("%d%d",&n,&m);
         for(i=;i<n;i++)
             scanf("%d",&aa[i]);
         for(i=;i<m;i++)
             scanf("%d",&bb[i]);
         printf("%d\n",exd_kmp(aa,bb,n,m));
     }
     return ;
 }

java代码：

 import java.util.Scanner;

 public class Main {

     public static void main(String args [])
     {
         mt aa = new mt();
        Scanner reader =new Scanner(System.in);
        int test=reader.nextInt();
        while((test--)>0)
        {
           aa.lena=reader.nextInt();
           aa.lenb=reader.nextInt();
           aa.init(aa.lena+1, aa.lenb+1);
           for(int i=0;i<aa.lena;i++)
             aa.a[i]=reader.nextInt();
           for(int i=0;i<aa.lenb;i++)
             aa.b[i]=reader.nextInt();
              kmp(aa);
        }
     }
     private static void kmp(mt aa)
     {
       int i,j;
       aa.next[i=0]=-1;
       j=-1;
       while(i<aa.lenb)
       {
         if(j==-1||aa.b[i]==aa.b[j])
         {
             i++;
             j++;
           if(aa.b[i]==aa.b[j])
              aa.next[i]=aa.next[j];
           else
              aa.next[i]=j;
         }
         else
             j=aa.next[j];
       }
       i=j=0;
       while(i<aa.lena&&j<aa.lenb)
       {
           if(j==-1||aa.a[i]==aa.b[j])
           {
              i++;
              j++;
           }
           else    j=aa.next[j];
       }
       if(j==aa.lenb)
           out(i-aa.lenb+1);
       else
          out(-1);
     }
     private static void out(int aa)
     {
         System.out.println(aa);
     }
 }
 class mt
 {
     int [] b,a,next;
     int  lena,lenb;
     void init(int lena,int lenb)
     {
            a=new int [lena];
            b=new int [lenb];
        next =new int [lenb];
     }
 }