Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5898    Accepted Submission(s): 2652

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  

Recommend lcy      

#include<stdio.h>
int a[1000010],b[10010];
int next[10010];
int n,m;
void getNext()
{
    int j,k;
    j=0;
    k=-1;
    next[0]=-1;
    while(j<m)
    {
        if(k==-1||b[j]==b[k])
          next[++j]=++k;
        else k=next[k];
    }    
}  
//返回首次出现的位置 
int KMP_Index()
{
    int i=0,j=0;
    getNext();
    
    while(i<n && j<m)
    {
        if(j==-1||a[i]==b[j])
        {
            i++;
            j++;
        }    
        else j=next[j];
        
    }    
    if(j==m) return i-m+1;
    else 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]);
        printf("%d\n",KMP_Index());
    }    
    return 0;
}    

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
using namespace std;
const int MAXN=1000010;

int a[MAXN];
int b[MAXN];

int n,m;
int next[MAXN];
/*
根据定义next[0]=-1，假设next[j]=k, 即P[0...k-1]==P[j-k,j-1]
1)若P[j]==P[k]，则有P[0..k]==P[j-k+1,j]，很显然，next[j+1]=next[j]+1=k+1;
2)若P[j]!=P[k]，则可以把其看做模式匹配的问题，即匹配失败的时候，k值如何移动，显然k=next[k]。
*/

void getNext()
{
    int j,k;
    next[0]=-1;
    j=0;
    k=-1;
    while(j<m)
    {
        if(k==-1||b[j]==b[k])
        {
            j++;
            k++;
            next[j]=k;
        }
        else k=next[k];
    }
}

int KMP_Index()
{
    int i=0,j=0;
    getNext();
    while(i<n&&j<m)
    {
        if(j==-1||a[i]==b[j])
        {
            i++;j++;
        }
        else j=next[j];
    }
    if(j==m)return i-m+1;
    else 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]);
        printf("%d\n",KMP_Index());
    }
    return 0;
}