Number Sequence
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 29376 Accepted Submission(s): 12350
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
题解:KMP算法的基本应用。如果对KMP还不是很了解,可以看这篇博文:点击打开链接
代码如下:
#include<cstdio>
#include<cstring>
int a[1000005],b[10005];
int Next[10005];
int n,m;
int pos;
void Get_Next(){
int i,k=0;
Next[0]=0;
for(i=1;i<m;i++){
while(k>0&&b[i]!=b[k]){
k=Next[k-1];
}
if(b[k]==b[i])
k++;
Next[i]=k;
}
}
void Kmp(){
int i,k=0;
for(i=0;i<n;i++){
while(k>0&&a[i]!=b[k]){
k=Next[k-1];
}
if(b[k]==a[i])
k++;
if(k==m){
pos=i-m+1;
break;
}
}
if(pos==-1) pos=-2;
}
int main()
{
int T;
scanf("%d",&T);
while(T--){
scanf("%d%d",&n,&m);
memset(Next,0,sizeof(Next));
for(int i=0;i<n;i++)
scanf("%d",&a[i]);
for(int i=0;i<m;i++)
scanf("%d",&b[i]);
pos=-1;
Get_Next();
Kmp();
printf("%d\n",pos+1);
}
return 0;
}