Number Sequence hdu 1711

题目链接：点击打开链接

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的模板题，kmp的难点就在于next数组，在写next数组时不要忘记可以把一个串看成两个，比如abcaabcd可以看成主串为abcd，模式串为abca

代码：

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int n,m;int a[1000005],b[10005],nextt[10005];void init(){    memset(a,0,sizeof(a));    memset(b,0,sizeof(b));    memset(nextt,-1,sizeof(nextt));}void get_next(int n){    int j=-1,i=0;    while(i<n)    {         if(j==-1||b[j]==b[i])         {             i++;             j++;             nextt[i]=j;         }         else            j=nextt[j];    }}int kmp(){    get_next(m);    int i=0,j=0;    while(i<n&&j<m)    {        if(j==-1||a[i]==b[j])        {            i++;            j++;        }        else            j=nextt[j];        if(j==m)            return i-j+1;    }    return -1;}int main(){    int t;    scanf("%d",&t);    while(t--)    {        init();        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]);        printf("%d\n",kmp());    }}