Number Sequence
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 24293 Accepted Submission(s): 10314
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
kmp 板子题
1 #include<iostream> 2 #include<cstdio> 3 #include<cmath> 4 #include<cstring> 5 #include<algorithm> 6 #include<queue> 7 #include<map> 8 #include<set> 9 #include<vector> 10 #include<cstdlib> 11 #include<string> 12 #define eps 0.000000001 13 typedef long long ll; 14 typedef unsigned long long LL; 15 using namespace std; 16 const int N=1000000+100; 17 int a[N],b[N]; 18 int m,n; 19 void get_next(int T[],int next[]){ 20 int j=0,k=-1; 21 next[0]=-1; 22 while(j<n){ 23 if(k==-1||T[j]==T[k]){ 24 j++; 25 k++; 26 next[j]=k; 27 //cout<<k<<" "; 28 } 29 else 30 k=next[k]; 31 } 32 } 33 void kmp(int S[],int T[]){ 34 int i=0,j=0; 35 int next[100000]; 36 get_next(T,next); 37 while(i<m){ 38 if(j==-1||S[i]==T[j]){ 39 i++; 40 j++; 41 } 42 else 43 j=next[j]; 44 if(j==n){ 45 printf("%d ",i-n+1); 46 return; 47 } 48 } 49 printf("-1 "); 50 } 51 int main(){ 52 int t; 53 scanf("%d",&t); 54 while(t--){ 55 scanf("%d%d",&m,&n); 56 int x,y; 57 for(int i=0;i<m;i++){ 58 scanf("%d",&a[i]); 59 } 60 for(int i=0;i<n;i++){ 61 scanf("%d",&b[i]); 62 } 63 //get_next(b); 64 kmp(a,b); 65 } 66 }