1711 Number Sequence(kmp)

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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

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代码:

#include <iostream>
#include <cstdio>
using namespace std;
int p[],s[];
int n,m;
int nex[]; void get()
{
int plen=m;
nex[]=-;
int k=-,j=;
while(j < plen){
if(k==- || p[j] == p[k]){
++j;
++k;
if(p[j] != p[k])
nex[j]=k;
else
nex[j]=nex[k];
}
else{
k=nex[k];
}
}
} int kmp()
{
int i=,j=;
int slen=n;
int plen=m;
while(i < slen && j< plen){
if(j==- || s[i]==p[j]){
++i;
++j;
}
else{
j=nex[j];
}
}
if(j == plen)
return i-j+;
else
return -;
} int main()
{
int t;
scanf("%d",&t);
while(t--){
scanf("%d%d",&n,&m);
for(int i=; i<n; i++)
scanf("%d",&s[i]);
for(int i=; i<m; i++)
scanf("%d",&p[i]);
get();
printf("%d\n",kmp());
}
}