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 |
http://acm.hdu.edu.cn/showproblem.php?pid=1711
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int a[1000001],n,m;
int b[1000001];
int Next[1000005];
void getnext()
{
/*int j=0,k=-1;
next[0]=-1;
while(j<m-1)
{
if(k==-1||b[j]==b[k])
{
j++;
k++;
next[j]=k;
}
else{
k=next[k];
}
}
*/
int j=0,k=-1;
Next[0]=-1;
while(j<m-1)
{
if(k==-1||b[j]==b[k])
{
j++;
k++;
Next[j]=k;
}
else
k=Next[k];
}
}
int KMP()
{
/*int i=0,j=0;
while(i<n&&j<m){
if(j==-1||a[i]==b[i]){
j++;
i++;
}else{
j=next[j];
}
}
if(j==m){
return i-j+1;
}
return -1;
*/
int i=0,j=0;
while(i<n&&j<m)
{
if(j==-1||a[i]==b[j])
{
i++;
j++;
}
else
j=Next[j];
}
if(j==m)
return i-j+1;//j是匹配位置;
return -1;
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%d %d",&n,&m);
for(int i=0; i<n; i++)
scanf("%d",&a[i]);
for(int i=0; i<m; i++)
scanf("%d",&b[i]);
getnext();
int ans=KMP();
printf("%d\n",ans);
}
return 0;
}