HDU 1711 Number Sequence(KMP入门)

A - Number Sequence

HDU - 1711

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

鹏哥口中短小精悍的KMP来了,寻找子字符串的最快算法,KMP精髓在于寻找next数组,这道题就是简单的两个函数调用就行

#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
int a[1000005], b[1000005];
int ne[1000005];
int aa, bb;
void getnext()//得到next数组
{
    int i = 0, j = -1;
    ne[0] = -1;
    while(i < bb)
    {
        if(j == -1||b[i] == b[j])
        ne[++i] = ++j;
        else j = ne[j];
    }
}
int kmpcompare()//判断是否存在位置符合条件
{
//    int len1 = strlen(a);
//    int len2 = strlen(b);
    getnext();
    int i = 0, j = 0;
    while(i < aa && j < bb)
    {
        if(j == -1|| a[i] == b[j]){j ++; i ++;}
        else j = ne[j];
    }
    if(j == bb)return i - j + 1;
    else return -1;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&aa,&bb);
        for(int i = 0; i < aa; i ++)
            scanf("%d",&a[i]);
        for(int i = 0; i < bb; i ++)
            scanf("%d",&b[i]);
        printf("%d\n",kmpcompare());
    }
    return 0;
}

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