HDU 1711 Number Sequence
| Time Limit: 5000MS | Memory Limit: 32768KB | 64bit IO Format: %I64d & %I64u |
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
思路：裸KMP
<span style="font-size:18px;">#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <queue>
#include <stack>
#include <set>
#include <vector>
#include<cstdlib>
#include <cmath>
#pragma comment (linker,"/STACK:102400000,102400000")
using namespace std;
#define maxn 1000500
int next_[maxn];
int str1[maxn];
int str2[maxn];
int len_1,len_2;
int conut;
void get_next()
{
int i=0;
int j=-1;
next_[i]=j;
while(i<len_2)
{
if(j==-1 || str2[i]==str2[j])
{
++i;
++j;
next_[i]=j;
}
else
j=next_[j];
}
}
void kmp()
{
get_next();
int i=0;
int j=0;
int len=len_1;
while(i<len)
{
if(j==-1 || str1[i]==str2[j])
{
++i;
++j;
}
else
j=next_[j];
if(j==len_2)
{
cout<<i-len_2+1<<endl;
return;
}
}
cout<<"-1"<<endl;
}
int main()
{
int T;
cin>>T;
while(T--)
{
scanf("%d%d",&len_1,&len_2);
memset(next_,-1,sizeof(next_));
conut=0;
for(int i=0;i<len_1;i++) scanf("%d",&str1[i]);
for(int i=0;i<len_2;i++) scanf("%d",&str2[i]);
kmp();
//cout<<conut<<endl;
}
return 0;
}</span>
