Number Sequence
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
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
The main idea of the topic: You enter two numbers. Find the subsequence of the first sequence to match the subscript of the first number in the second sequence.
Analysis: KMP directly returns the subscript when matched
If you don’t understand the KMP algorithm, please learn from Baidu. My template is relatively simple.
#include "iostream"
#include "cstdio"
#include "cstring"
#include "algorithm"
using namespace std;
const int maxn = 1e6+5;
int NEXT_A[maxn];
int num_a[maxn];
int num_b[maxn];
int n;
int m;
void MakeNext( int *num ){
int i = 0;
int j = -1;
NEXT_A[0] = -1;
while( i < m ){
if( j == -1 || num[j] == num[i] ){
NEXT_A[++i] = ++j;
}
else{
j = NEXT_A[j];
}
}
}
int kmp(){
int i = 0;
int j = -1;
int cnt = 0;
while( i<n && j<m ){
if( j == -1 || num_a[i] == num_b[j] ){
i++;
j++;
}
else{
j = NEXT_A[j];
}
if( j == m ){
cnt = i;
cnt -= (m-1);
return cnt;
}
}
return -1;
}
int main(){
int T;
bool flag;
scanf("%d",&T);
while( T-- ){
flag = true;
scanf("%d%d",&n,&m);
for( int i=0 ; i<n ; i++ ) scanf("%d",&num_a[i]);
for( int i=0 ; i<m ; i++ ) scanf("%d",&num_b[i]);
MakeNext(num_b);
for( int i=0 ; i<m ; i++ ){
if( num_a[i] != num_b[i] ){
flag = false;
break;
}
}
if( flag ){
puts("1");
}
else{
int ans = kmp();
printf("%d\n",ans);
}
}
return 0;
}