hdoj 1711 Number Sequence 【KMP】

Number Sequence

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

Total Submission(s): 15101 Accepted Submission(s): 6623

Problem Description

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<stdio.h>
#include<string.h>
#include<stdlib.h>
const int N=1000000+10;
const int M=10000+10;
int a
,b[M];
int next
;
int ls,lm;
void getn()
{
int i = 1,j = 0;
next[0] = -1;
next[1] = 0;
while(i < ls)
{
if(j == -1||b[i] == b[j])
{
i++,j++;
next[i] = j;  //求出 子串 的模式串
}
else
j = next[j];
}
}
int kmp()
{

int i,j;
getn();
i = 0;j = 0;
while(i < lm&&j != ls)
{
if(j == -1||a[i]==b[j]) //逐位比较子串与母串
{
i++,j++;
}
else
j = next[j];
}
if( j == ls )
return i-ls+1; //从0开始所以要再加上 1.
else
return -1; //不满足的话 输出 -1.
}

int main()
{
int n,i,j,k;
scanf("%d",&n);
while(n--)
{
scanf("%d%d",&lm,&ls);
for(i = 0;i < lm; i++)
scanf("%d",&a[i]);
for(j = 0;j < ls;j++)
scanf("%d",&b[j]);
k = kmp();
printf("%d\n",k);
}
return 0;
}