HDU 1711 Number Sequence【数字KMP】

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Number Sequence

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 22886 Accepted Submission(s): 9781

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

原题链接：http://acm.hdu.edu.cn/showproblem.php?pid=1711

数字的KMP。

第二个数组中的数字在第一个数组中出现的位置

AC代码：

#include <iostream>#include <cstring>#include <cstdio>using namespace std;const int maxn= 1000000+5;const int maxm= 10000+5;int lena,lenb;int a[maxn];int b[maxm];int nextt[maxm];void getNext(){    int i=0,j=-1;    nextt[0]=-1;//不要忘记呀！！！！    while(i<lenb)    {        if(j==-1||b[i]==b[j])        {            i++,j++;            nextt[i]=j;        }        else            j=nextt[j];    }}int KMP(){    int i=0,j=0;    getNext();    while(i<lena)    {        if(j==-1||a[i]==b[j])            i++,j++;        else            j=nextt[j];        if(j==lenb)            return i-j+1;    }    return -1;}int main(){    int T;    //freopen("data/1711.txt","r",stdin);    cin>>T;    while(T--)    {        cin>>lena>>lenb;        for(int i=0;i<lena;i++)            cin>>a[i];        for(int i=0;i<lenb;i++)            cin>>b[i];        memset(nextt,0,sizeof(nextt));        cout<<KMP()<<endl;    }    return 0;}

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