HDOJ ---1711 Number Sequence

时间：2014-04-29 11:28:46 收藏：0 阅读：354

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

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

-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 }

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