HDU--杭电--1711--Number Sequence--KMP--水题，看毛片算法基础死方法运用

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8126    Accepted Submission(s): 3688

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

 

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int next[11111],s1[1111111],s2[11111];
void getnext(int s[],int l)  //求next数组
{
    int i=0,j=-1;next[i]=j;
    while(i<l)
        if(j==-1||s[i]==s[j])next[++i]=++j;
        else j=next[j];
}
int KMP(int l1,int l2)  //匹配
{
    int i=0,j=0;
    while(i<l1&&j<l2)
        if(j==-1||s1[i]==s2[j])i++,j++;
        else j=next[j];
    if(j==l2)return i-l2+1;
    return 0;
}
int main (void)
{
    int t,i,n,m;
    cin>>t;
    while(t--&&scanf("%d%d",&n,&m))
    {
        for(i=0;i<n;i++)
            scanf("%d",&s1[i]);
        for(i=0;i<m;i++)
            scanf("%d",&s2[i]);
        getnext(s2,m);i=KMP(n,m);
        if(i)printf("%d\n",i);
        else printf("-1\n");
    }
    return 0;
}