Number Sequence

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 54826 Accepted Submission(s): 21988

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

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

Source

HDU 2007-Spring Programming Contest

Recommend

lcy

题目意思：就是首先输入t表示要处理数据的组数，然后输入n和m，后面跟着两个长度为n和m的数组。找到长度为m的数组在长度为n数组里面的出现位置（位置要最小）。

题目思路：只需要在输入的地方稍微处理一下（原先的字符串输入改为数字输入）就行，然后直接kmp。

ac代码：

#include<stdio.h>
#include<string>
const int maxn=1e6+10;
const int maxm=1e4+10;
int t,n,m,s[maxn],p[maxn];
int next[maxm];
void GetNext()
{
    int plen=0;
    int slen=-1;
    next[0]=-1;
    while(plen<m)
    {
        if(slen==-1||p[plen]==p[slen])
        {
            plen++;slen++;
            if(p[plen]!=p[slen])next[plen]=slen;
            else next[plen]=next[slen];
        }
        else slen=next[slen];
    }
}
int kmp()
{
    int plen=0;
    int slen=0;
    while(plen<m&&slen<n)
    {
        if(plen==-1||p[plen]==s[slen])
        {
            plen++;slen++;
        }
        else plen=next[plen];
    }
    if(plen==m)
    return slen-plen+1;
    else return -1;
}
int main()
{
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        for(int i=0;i<n;i++)scanf("%d",&s[i]);
        for(int i=0;i<m;i++)scanf("%d",&p[i]);
        GetNext();
        printf("%d
",kmp());
    }
    return 0;
}