hdu1005 Number Sequence(kmp字符串比较)

Given two sequences of numbers : a[1], a[2], …… , a
, 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
. 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<cstdio>
#include<cstring>
#include<iostream>
using namespace std;

int ne[10010];
int a[1000010], b[10010];

void GetNext(int lenb)
{
int i = 0, j = -1;
ne[0] = -1;
while(i < lenb)
{
if(j == -1 || b[i] == b[j]){
ne[++i] = ++j;
}
else j = ne[j];
}
}

int KmpCompare(int lena, int lenb)
{
int i = 0, j = 0;
while(i < lena && j < lenb)
{
if(j == -1 || a[i] == b[j]){
i++;
j++;
if(j == lenb) return i - j + 1;
}
else j = ne[j];
}
return -1;
}

int main()
{
int t, c, d, e, i;
scanf("%d", &t);
while(t--)
{
scanf("%d %d", &c, &d);
for(i=0; i<c; i++){
scanf("%d", &a[i]);
}
for(i=0; i<d; i++){
scanf("%d", &b[i]);
}
GetNext(d);
e = KmpCompare(c, d);
printf("%d\n", e);
}
return 0;
}