HDU 1711 Number Sequence
Time Limit: 5000MS | Memory Limit: 32768KB | 64bit IO Format: %I64d & %I64u |
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
思路:裸KMP
<span style="font-size:18px;">#include <iostream> #include <algorithm> #include <cstdio> #include <cstring> #include <queue> #include <stack> #include <set> #include <vector> #include<cstdlib> #include <cmath> #pragma comment (linker,"/STACK:102400000,102400000") using namespace std; #define maxn 1000500 int next_[maxn]; int str1[maxn]; int str2[maxn]; int len_1,len_2; int conut; void get_next() { int i=0; int j=-1; next_[i]=j; while(i<len_2) { if(j==-1 || str2[i]==str2[j]) { ++i; ++j; next_[i]=j; } else j=next_[j]; } } void kmp() { get_next(); int i=0; int j=0; int len=len_1; while(i<len) { if(j==-1 || str1[i]==str2[j]) { ++i; ++j; } else j=next_[j]; if(j==len_2) { cout<<i-len_2+1<<endl; return; } } cout<<"-1"<<endl; } int main() { int T; cin>>T; while(T--) { scanf("%d%d",&len_1,&len_2); memset(next_,-1,sizeof(next_)); conut=0; for(int i=0;i<len_1;i++) scanf("%d",&str1[i]); for(int i=0;i<len_2;i++) scanf("%d",&str2[i]); kmp(); //cout<<conut<<endl; } return 0; }</span>