HDU 1711 Number Sequence（KMP）

Number Sequence

Time Limit : 10000/5000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)
Total Submission(s) : 27 Accepted Submission(s) : 13
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


Source
HDU 2007-Spring Programming Contest

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxx=1e6+10;
int Next[maxx];
int s[maxx],p[maxx];
int len1,len2;
void getnext()
{
    int i=0,j=-1;
    Next[0]=-1;
    while(i<len1)
    {
        if(j==-1||p[i]==p[j])
        {
            ++i;
            ++j;
            if(p[i]==p[j])
                Next[i]=Next[j];
            else
                Next[i]=j;
        }
        else
            j=Next[j];
    }
}
int getkmp()
{
    int num=0,i=0,j=0;
    while(i<len2)
    {
        if(j==-1|s[i]==p[j])
        {
            ++i;++j;
            if(j==len1)
                return i-len1+1;
        }
        else
            j=Next[j];

    }
    return -1;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&len2,&len1);
        for(int i=0;i<len2;i++)
            scanf("%d",&s[i]);
         for(int i=0;i<len1;i++)
            scanf("%d",&p[i]);
        getnext();
        if(len2<len1)
            printf("-1
");
        else
            printf("%d
",getkmp());
    }
    return 0;
}