hdu1711 Number Sequence（KMP算法）

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 25812    Accepted Submission(s): 10909

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

 

Recommend

lcy   |   We have carefully selected several similar problems for you:  1358 3336 1686 3746 1251 

 

Statistic | Submit | Discuss | Note

直接套模版即可

#include <stdio.h>int a[1000000+5];int b[10000+5];int next[10000+5];int n,m;void setnext(){    int i=-1;    int j=0;    next[0]=-1;    while(j<m)    {        if(i==-1||a[i]==b[j])        {            i++;            j++;            next[j]=i;        }        else        {            i=next[i];        }    }}int kmp(){    int res=-1;    setnext();    int i=0,j=0;    while(i<n)    {        if(j==-1||a[i]==b[j])        {            i++;            j++;        }        else        {            j=next[j];        }        if(j==m)        {            res=i;            break;        }    }    if(res==-1)    return -1;    else    return res-m+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 j=0;j<m;j++)        {            scanf("%d",&b[j]);        }        printf("%d\n",kmp());    }    return 0;}