HDU 1711 Number Sequence 【KMP算法】

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11847    Accepted Submission(s): 5403

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

 

/*    KMP算法 */ #include<cstdio>#include<cstring>int n,m,s[1000002],p[10002],next[10002];void get_next(int* p){int i,j;i=0;next[0]=j=-1;while(i<m-1){if(j==-1||p[i]==p[j]){i++;j++;next[i]=j;}else{j=next[j];}}}int kmp(int* s, int* p){int i,j;i=j=0;while(i<n){if(j==-1||s[i]==p[j]){i++,j++;}else j=next[j];if(j==m){return i-m+1;}}return -1; }int main(){int T;scanf("%d",&T);while(T--){scanf("%d %d",&n,&m);for(int i=0; i<n; i++)scanf("%d",&s[i]);for(int i=0; i<m; i++)scanf("%d",&p[i]);get_next(p);printf("%d\n",kmp(s,p));}return 0;}