hdoj-1711Number Sequence(数组Kmp)

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 31863    Accepted Submission(s): 13380

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

 

题目链接

就是把字符转换成数组比较，kmp的一个应用。

#include <iostream>#include <cstdio>#include <cstdlib>#include <algorithm>using namespace std;const int maxn=1e6+5;int pl,sl;int s[maxn],p[maxn];int next1[maxn];int sum;void next(){    int k=-1,j=0;    next1[0]=-1;    while(j<pl)    {        if(k==-1||p[j]==p[k])        {            j++;            k++;            next1[j]=k;        }        else        k=next1[k];    }}int KMP(){    int i=0,j=0;    next();    while(j<pl&&i<sl)    {            if(j==-1||s[i]==p[j])            {                i++;j++;            }            else j=next1[j];            if(j==pl)            {                sum=i-pl+1;j=next1[j];                break;            }    }    return sum;}int main(){    int t;    scanf("%d",&t);    while(t--)    {        sum=-1;        scanf("%d%d",&sl,&pl);        for(int i=0;i<sl;i++)        scanf("%d",&s[i]);        for(int i=0;i<pl;i++)        scanf("%d",&p[i]);        int ans=KMP();        cout<<ans<<endl;    }    return 0;}