HDU1711-Number Sequence

Number Sequence

                                                                       Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
                                                                                                  Total Submission(s): 22445    Accepted Submission(s): 9596

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

 

Source

HDU 2007-Spring Programming Contest

#include <iostream>#include <stdio.h>#include <queue>#include <string.h>using namespace std;int a[1000009],b[10009],nt[10009];int n,m;void get_next(){    nt[0]=-1;    for(int i=0; i<m; i++)    {        int k=nt[i];        while(k>=0&&b[i]!=b[k])            k=nt[k];        nt[i+1]=k+1;    }}int kmp(){    int i=0,j=0;    while(i<n)    {        if(a[i]==b[j]||j==-1)            i++,j++;        else            j=nt[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",&a[i]);        for(int i=0; i<m; i++)            scanf("%d",&b[i]);        get_next();        printf("%d\n",kmp());    }    return 0;}