HDU1711 Number Sequence(KMP模板题)
Number Sequence
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 40053 Accepted Submission(s): 16510
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
#include<iostream>
#include<string.h>
using namespace std;
int tt,pp;
int t[1000011],p[11111],next1[11111];
void getnext()
{
int i=0,j=-1;
next1[0]=-1;
while(i<pp-1)
{
if(j==-1||p[i]==p[j])
next1[++i]=++j;
else
j=next1[j];
}
return;
}
int kmp()
{
int i=0,j=0;
while(i<tt)
{
if(j==-1||t[i]==p[j])
i++,j++;
else
j=next1[j];
if(j==pp)
return i-j+1;
}
return -1;
}
int main()
{
int n;
scanf("%d",&n);
while(n--)
{
scanf("%d%d",&tt,&pp);
for(int i=0;i<tt;i++)
scanf("%d",&t[i]);
for(int i=0;i<pp;i++)
scanf("%d",&p[i]);
getnext();
printf("%d\n",kmp());
}
return 0;
}
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