Number Sequence
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 15052 Accepted Submission(s): 6597
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
Source
KMP模板题,返回模式串第一次出现在主串中的位置
#pragma comment(linker, "/STACK:1024000000,1024000000") #include<cstdio> #include<string> #include<iostream> #include<cstring> #include<cmath> #include<stack> #include<queue> #include<vector> #include<map> #include<stdlib.h> #include<algorithm> #define LL __int64 using namespace std; const int MAXN=1000000+5; int S[MAXN],T[MAXN]; int nex[MAXN]; int n,m; void KMP_nex() { int j,k; j=0;k=-1,nex[0]=-1; while(j<m) { if(k==-1 || T[j]==T[k]) nex[++j]=++k; else k=nex[k]; } } int KMP_ID() { int i=0,j=0; KMP_nex(); while(i<n && j<m) { if(j==-1 || S[i]==T[j]) i++,j++; else j=nex[j]; } if(j==m) return i-m+1; else return -1; } int main() { //freopen("in.txt","r",stdin); int kase; scanf("%d",&kase); while(kase--) { scanf("%d %d",&n,&m); for(int i=0;i<n;i++) scanf("%d",&S[i]); for(int i=0;i<m;i++) scanf("%d",&T[i]); printf("%d ",KMP_ID()); } return 0; }