Number Sequence
Time Limit: 10000/5000 MS
(Java/Others) Memory Limit: 32768/32768 K
(Java/Others)
Total Submission(s): 9899 Accepted
Submission(s): 4518
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:
1 #include<cstdio>
2 #define MAXN 1000010
3 int T[MAXN], P[MAXN/10], fail[MAXN/10], N, M, t;
4 void getFail(){
5 fail[0] = -1;
6 for(int i = 1, j = -1;i < M;i ++){
7 while(P[i] != P[j+1] && j >= 0) j = fail[j];
8 if(P[i] == P[j+1]) j++;
9 fail[i] = j;
10 }
11 }
12 void Kmp(){
13 getFail();
14 for(int i = 0, j = 0;i < N;i ++){
15 while(T[i] != P[j] && j) j = fail[j-1] + 1;
16 if(T[i] == P[j]) j++;
17 if(j == M){
18 printf("%d\n", i-M+2);
19 return;
20 }
21 }
22 printf("-1\n");
23 }
24 int main(){
25 scanf("%d", &t);
26 while(t--){
27 scanf("%d%d", &N, &M);
28 for(int i = 0;i < N;i ++) scanf("%d", T+i);
29 for(int i = 0;i < M;i ++) scanf("%d", P+i);
30 Kmp();
31 }
32 return 0;
33 }
HDOJ ---1711 Number Sequence,码迷,mamicode.com