HDU 1711

简单的KMP

好吧好吧

还是花了一上午

next[0] =-1

还是看注释吧

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

 

#include <iostream>
using namespace std;

int T[1000005];//不明白自己动态分配的话就会ACCESS_VIOLENCE
int P[10005];//不管了以后都写外面好了
int next[10005];

int main()
{
    int t;
    cin >> t;
    while (t--)
    {
        int M, N;
        cin >> N >> M;

        for (int i = 0; i < N; i++)
        {
            cin >> T[i];
        }

        for (int i = 0; i < M; i++)
        {
            cin >> P[i];
        }

        for (int i = 0; i < 10005; i++)
        {
            next[i] = 0;
        }

        for (int i = 1; i < M; i++)//获得next数组
        {
            int j = next[i];
            while (j&&P[i] != P[j])//如果不相同就循环一直到j=0或者找到j使得P[i]=P[j];
                j = next[j];
            next[i + 1] = (P[i] == P[j] ? j + 1 : 0);//递归获得next
        }
        // next[0] 设置为 -1
        next[0] = -1;
        int ans = -1;
        int j = 0;
        if (M == 1)//m长为1时
        {
            for (int i = 0; i < N; i++)
            {
                if (T[i] == P[j])
                {
                    ans = i + 1;
                    break;
                }
            }
        }
        else
        {
            for (int i = 0; i<N; i++)
            {
                //     j > -1 , 保证完全匹配失败的时候，j能返回到0
                while (j > -1 && T[i] != P[j])
                    j = next[j];
                j++;
                if (j == M) {
                    ans = i - M + 1 + 1;
                    break;
                }
            }
        }

        cout << ans << endl;
    }
    return 0;
}