HDU-1711-Number Sequence

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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

  
   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

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;
}