(HDU

(HDU - 1711)Number Sequence

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 30625 Accepted Submission(s): 12882

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

2
13 5
1 2 1 2 3 1 2 3 1 3 2 1 2
1 2 3 1 3
13 5
1 2 1 2 3 1 2 3 1 3 2 1 2
1 2 3 2 1

Sample Output

6
-1

题目大意：给出两个数字串a,b问在a串中能否找到b串，若能找到就输出第一个找到的第一个字符的位置，如果不能就输出-1。

思路：kmp模板题。

#include<cstdio>using namespace std;int a[1000005],b[10005],Next[10005];int n,m;void getNext(){    Next[0]=Next[1]=0;    for(int i=1;i<m;i++)    {        int j=Next[i];        while(j&&b[i]!=b[j]) j=Next[j];        if(b[i]==b[j]) Next[i+1]=j+1;        else Next[i+1]=0;    }}int kmp(){    getNext();    int j=0;    for(int i=0;i<n;i++)    {        while(j&&a[i]!=b[j]) j=Next[j];        if(a[i]==b[j]) j++;        if(j==m) return i-m+2;     }    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);        int ans=kmp();        printf("%d\n",ans);    }    return 0;}