HDU-1711-Number Sequence

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 29338    Accepted Submission(s): 12332

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

213 51 2 1 2 3 1 2 3 1 3 2 1 21 2 3 1 313 51 2 1 2 3 1 2 3 1 3 2 1 21 2 3 2 1

 

Sample Output

6-1

 

Source

HDU 2007-Spring Programming Contest

这道题是KMP的模板题，也没什么可所说的；

AC代码：

#include<cstdio>#include<cstdlib>#include<algorithm>using namespace std;const  int maxn=1e6+10;int a[maxn],b[maxn];int p[1000000];int Kmp(int* a,int n,int* b,int m){int i=0,j=0;while(i<n){if(j==-1||a[i]==b[j]){++i;++j;if(j==m){return i-m+1;}}else{j=p[j];}}return -1;}void getnext(int *b,int m)//求next数组； {int i=0,j=0;p[0]=-1;j=p[i];while(i<m){if(j==-1||b[i]==b[j]){p[++i]=++j;}else{j=p[j];}}}int main(){int T,n,m;scanf("%d",&T);while(T--){scanf("%d %d",&n,&m);for(int i=0;i<n;i++){scanf("%d",&a[i]);}for(int j=0;j<m;j++){scanf("%d",&b[j]);}getnext(b,m);printf("%d\n",Kmp(a,n,b,m));}return 0;}