hdu1711 （Number Sequence） KMP模板

[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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].

[align=left]Output[/align]
For each test case, you should output one line which only contain K described above. If no such K exists, output -1 instead.

[align=left]Sample Input[/align]

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

[align=left]Sample Output[/align]

6
-1

KMP模板：

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
#define N 1000006
#define M 10004
int a
,b[M];
int n,m;
int next[M];
int t;
void Get_next()
{
int i=0,j=-1;
next[0]=-1;
while(i<m)
{
if(j==-1||b[i]==b[j])
{
i++;
j++;
next[i]=j;
}
else
j=next[j];
}
}
int KMP()
{
Get_next();
int i=0,j=0;
while(i<n&&j<m)
{
if(j==-1||a[i]==b[j])
{
i++;
j++;
}
else
j=next[j];
}
if(j==m)
return i-m+1;
else
return -1;
}

int main()
{
scanf("%d",&t);
while(t--)
{
memset(a,0,sizeof(a));
memset(b,0,sizeof(b));
memset(next,-1,sizeof(next));
scanf("%d%d",&n,&m);
int i,j;
for(i=0;i<n;i++)
scanf("%d",&a[i]);
for(j=0;j<m;j++)
scanf("%d",&b[j]);
printf("%d\n",KMP());
}
return 0;
}