HDU_1711_NumberSequence

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 18002    Accepted Submission(s): 7874

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

 

Source

HDU 2007-Spring Programming Contest

 

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KMP的裸题

#include <iostream>#include <stdio.h>#include <string>#include <string.h>using namespace std;const int MN=1e6+5;const int MM=1e4+5;int s1[MN],s2[MM];int nex[MM];int f;void getnext(int m){    int j=0,k=-1;    nex[0]=-1;    while(j<m)    {        if(k==-1||s2[k]==s2[j])            nex[++j]=++k;        else            k=nex[k];    }}int getp(int m,int n){    int j=0,k=0;    while(j<n)    {        if(k==-1||s2[k]==s1[j])        {            k++;j++;        }        else            k=nex[k];        if(k==m)            return j-m+1;    }    return -1;}int main(){    int t;    int n,m;    scanf("%d",&t);    while(t--)    {        f=-1;        memset(nex,0,sizeof(nex));        scanf("%d%d",&n,&m);        for(int i=0;i<n;i++)            scanf("%d",&s1[i]);        for(int i=0;i<m;i++)            scanf("%d",&s2[i]);        getnext(m);        //for(int i=0;i<10;i++)        //   cout<<i<<" "<<nex[i]<<endl;        printf("%d\n",getp(m,n));    }    return 0;}