kmp入门---hd1711

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 29786    Accepted Submission(s): 12548

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

一直定义char型的s【】  a【】  一直wr  现在换成int 就过了  真是大意啊

#include<iostream>#include<stdio.h>using namespace std;#include<string>#include<string.h>const int N=1000010;int  s[N],a[N];int t,i,n,m,next1[N];int j,k;void init(){    cin>>n>>m;    for(i=0;i<n;i++)    scanf("%d",&s[i]);    for(i=0;i<m;i++)    scanf("%d",&a[i]);}void get_next(int  *a,int *next1){    k=-1,j=0;    next1[0]=-1;    while(j<m){        if(k==-1||a[k]==a[j]){            k++;            j++;            if(a[k]!=a[j])    next1[j]=k;            else    next1[j]=next1[k];        }        else    k=next1[k];    }}int kmp(int  *s,int  *a){    get_next(a,next1);    i=j=0;    while(i<n){        if(j==-1||a[j]==s[i])i++,j++;        else    j=next1[j];        if(j==m)            return i-j+1;    }    return -1;}int main(){    scanf("%d",&t);    while(t--)    {        init();        cout<<kmp(s,a)<<endl;    }    return 0;}