hdu1711Number Sequence（Kmp）

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8321    Accepted Submission(s): 3795

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.

#include <iostream>#include <string.h>#include <string>#include <cstdio>#include <algorithm>#define MAXN 1000001#define MAX 100011using namespace std;int len1, len2;int p[MAXN];int str1[MAXN], str2[MAX];void getp(){    p[0] = -1;    for (int i = 1, j = -1; i < len2; i++)    {        while (j >= 0 && str2[i] != str2[j + 1])        {            j = p[j];        }        if (str2[i] == str2[j + 1])        {            j += 1;        }        p[i] = j;    }}bool kmp(){    int num = 1;    for (int i = 0, j = -1; i < len1; i++)    {        if (str1[i] != str2[j + 1])        {            while (j >= 0 && str1[i] != str2[j + 1])            {                j = p[j];            }            num++;        }        if (str1[i] == str2[j + 1])        {            j += 1;        }        if (j == len2 - 1)        {            cout << i - j + 1 << endl;            return true;        }    }    cout << -1 << endl;    return false;}void input(){    int t;    cin >> t;    while (t--)    {        cin >> len1 >> len2;        for (int i = 0; i < len1; i++)        {            scanf("%d", &str1[i]);        }        for (int i = 0; i < len2; i++)        {            scanf("%d", &str2[i]);        }        getp();        kmp();    }}int main(){    input();    return 0;}