hdu 1711 Number Sequence

Number Sequence

Time Limit: 5000MS

Memory Limit: 32768KB

64bit IO Format: %I64d & %I64u

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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 
     

 

裸的KMP，输入时处理一下

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <stack>
#include <cmath>
#include <queue>
#include <map>
using namespace std;
#define N 2001000
int s[N];
int pattern[N];
int nest[N];
int n,m;
void getnext()
{
    //int len=strlen(pattern);
    int j=-1;
    nest[0]=-1;
    for(int i=1;i<m;i++) {
        while(j!=-1&&pattern[i]!=pattern[j+1])
            j=nest[j];
        if(pattern[j+1]==pattern[i])
            j++;
        nest[i]=j;
    }

}
int main()
{
    int T;
    scanf("%d",&T);
    //getchar();
    while(T--) {

        char ch;
        scanf("%d%d",&n,&m);
        //getchar();
        int i=0;
        for(int i=0;i<n;i++)
            scanf("%d",&s[i]);
        for(int i=0;i<m;i++)
            scanf("%d",&pattern[i]);
        if(m>n) {
            printf("-1");
            continue;
        }
        getnext();
        //int len1=strlen(s),len2=strlen(pattern);
        int j=-1;
      //  cout<<s<<endl<<pattern<<endl;
        for(i=0;i<n;i++) {
            while(j!=-1&&s[i]!=pattern[j+1])
                j=nest[j];
            if(s[i]==pattern[j+1])
                j++;
            if(j==m-1)
                break;
        }
        if(j!=m-1)
            printf("-1\n");
        else
            printf("%d\n",i-j+1);
    }
    return 0;
}