题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1711

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

Problem Description
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

题目大意:给出两个数字串:文本串和模式串,求模式串在文本串中第一次出现的位置。

解题思路:依然是KMP模板题……

 #include<cstdio>
#include<cstring>
#define MAXpat 10000+5
#define MAXstr 1000000+5
int len1,len2;
int Next[MAXpat],str[MAXstr],pat[MAXpat];
void getNext()
{
int i=, j=-;
Next[]=-;
while(i<len2)
{
if(j == - || pat[i] == pat[j]) Next[++i]=++j;
else j=Next[j];
}
}
int kmp()
{
getNext();
int i=, j=;
while(i<len1)
{
if(j == - || str[i] == pat[j]) i++, j++;
else j=Next[j];
if(j == len2) return i-j+;
}
return -;
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&len1,&len2);
for(int i=;i<len1;i++) scanf("%d",&str[i]);
for(int i=;i<len2;i++) scanf("%d",&pat[i]);
printf("%d\n",kmp());
}
}

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