Number Sequence（KMP之子串首次出现的位置）

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 30033    Accepted Submission(s): 12666


[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题意：找出第二个数组在第一个数组中第一次出现的位置,位置从1开始；

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int a[1000001],n,m;
int b[1000001];
int Next[1000005];
void getnext()
{
int j=0,k=-1;
Next[0]=-1;
while(j<m-1)
{
if(k==-1||b[j]==b[k])
{
j++;
k++;
Next[j]=k;
}
else
k=Next[k];
}
}
int KMP()
{
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-j+1;
return -1;
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%d %d",&n,&m);
for(int i=0; i<n; i++)
scanf("%d",&a[i]);
for(int i=0; i<m; i++)
scanf("%d",&b[i]);
getnext();
int ans=KMP();
printf("%d\n",ans);
}
return 0;
}