数据结构之KMP算法---hdu---Number Sequence
发布日期:2022-02-02 02:58:08 浏览次数:4 分类:技术文章

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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
213 51 2 1 2 3 1 2 3 1 3 2 1 21 2 3 1 313 51 2 1 2 3 1 2 3 1 3 2 1 21 2 3 2 1
 

Sample Output
6-1代码实现:   
   #include 
    
     #include 
     
      using namespace std;int next[10005];int a[1000005],b[10005];void getNext(int *b,int m,int n){
      
int i=0,j=-1;
next[0]=-1;
while(i
{
if(j==-1||b[i]==b[j])
{
i++;j++;next[i]=j;
}
else
{
j=next[j];
}
}}int main(){
int t,m,n;
cin>>t;
while(t--)
{
cin>>m>>n;
for(int i=0;i
{
cin>>a[i];
}
for(int j=0;j
{
cin>>b[j];
}
getNext(b,m,n);
int i=0,j=0,flag=0;
while(i
{
   if(j==-1||a[i]==b[j])
   {
   i++;j++;
   }
   else
   {
   j=next[j];
   }
   if(j==n)
   {
   flag++;
   cout< <
   break;
   }
}
if(flag==0)cout<<"-1"<
}
return 0;}

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[***.249.68.14]2022年05月24日 06时48分00秒

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