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Number Sequence（杭电1711）（KMP）
Number Sequence
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 11925 Accepted Submission(s): 5440
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
Source
HDU 2007-Spring Programming Contest
/*KMP算法关键是求next函数，第一道KMP题，最基础的题，
借鉴普浩冉的代码和讲解，还算有点理解吧= =
*/
#include<stdio.h>
int a[1000002];
int s[10002],next[10002];
int m,n;
void get_next(int *s)
{
int i=0,j=-1;
next[0]=-1;
while(i<m-1){
if(j==-1||s[i]==s[j])
{
i++;
j++;
next[i]=j;
}
else
j=next[j];
}
}
int kmp(int *a,int *s)
{
int i=0,j=0;
get_next(s);
while(i<=n-1)
{
if(j==-1||a[i]==s[j])
{
i++;
j++;
}
else
j=next[j];
if(j==m)
return i-m+1;
}
return -1;
}
int main()
{
int test,i;
scanf("%d",&test);
while(test--)
{
scanf("%d %d",&n,&m);
for(i=0;i<n;i++)
scanf("%d",&a[i]);
for(i=0;i<m;i++)
scanf("%d",&s[i]);
printf("%d\n",kmp(a,s));
}
return 0;
}Number Sequence（杭电1711）（KMP）
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