Number Sequence
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 11606 Accepted Submission(s): 5294
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
Mean:
给你s1,s2两个串，让你找到s2在s1中出现的第一个位置。
analyse:
KMP字符串水题.
Time complexity：O(n+m)
Source code：
#include<cstdio> #include<cstring> int l1 , l2; int a [ 1000010 ],b [ 10010 ], Next [ 10010 ]; void getNext() { Next [ 0 ] = 0; int i , k; for( i = 1 , k = 0; i < l2; ++ i) { while(b [ i ] !=b [ k ] && k > 0) k = Next [ k - 1 ]; if(b [ i ] ==b [ k ]) ++ k; Next [ i ] = k; } } int kmp() { getNext(); for( int i = 0 , k = 0; i < l1; ++ i) { while( a [ i ] !=b [ k ] && k > 0) k = Next [ k - 1 ]; if( a [ i ] ==b [ k ]) ++ k; if( k == l2) return i - l2 + 2; } return - 1; } int main() { int t; scanf( "%d" , & t); while( t --) { scanf( "%d %d" , & l1 , & l2); for( int i = 0; i < l1; ++ i) scanf( "%d" , & a [ i ]); for( int i = 0; i < l2; ++ i) scanf( "%d" , &b [ i ]); printf( "%d \n " , kmp()); } return 0; }