Number Sequence
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 16335 Accepted Submission(s):
7198
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的初级算法, 较简单理解吧
#include<stdio.h> #include<string.h> const int MAXN = 1e6+7; int mumStr[MAXN], sonStr[MAXN]; int Next[MAXN]; void GetNext(int s[], int N, int next[]) { int i=0, j=-1; next[0] = -1; while(i < N) { if(j==-1 || s[i]==s[j]) next[++i] = ++j; else j = next[j]; } } int KMP(int mumStr[], int sonStr[], int Mn, int Sn) { int i=0, j=0; GetNext(sonStr, Sn, Next); while(i < Mn) { while(j==-1 || (mumStr[i]==sonStr[j]&&i<Mn&&j<Sn) ) i++, j++; if(j == Sn) return i-Sn+1; j = Next[j]; } return -1; } int main() { int T; scanf("%d", &T); while(T--) { int i, Mn, Sn; scanf("%d%d", &Mn, &Sn); for(i=0; i<Mn; i++) scanf("%d", &mumStr[i]); for(i=0; i<Sn; i++) scanf("%d", &sonStr[i]); printf("%d\n", KMP(mumStr, sonStr, Mn, Sn)); } return 0; }