hdu1711Number Sequence(KMP)

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 16453    Accepted Submission(s): 7245

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

 题目大意：给你两个数字序列a[]和b[],求b[]序列在a[]序列中最早发生匹配的位置，若不存在，输出-1.

简单的KMP,需要注意的是，由于是都是数字序列，每个数字范围不确定，这是不能当作字符来处理，所以a[]、b[]应定义为整型；

AC代码：

#include<stdio.h>#include<string.h>int a[1000100],b[10100];int p[10010];int n,m;void getp(){int i=0,j=-1;p[i]=j;while(i<m){if(j==-1||b[i]==b[j]){i++,j++;p[i]=j;}else j=p[j];}}int Kmp(){int i=0,j=0,k=-1;;getp();while(i<n){if(j==-1||a[i]==b[j]){i++,j++;if(j==m){k=i;break;} }else j=p[j];}return k;}int main(){int t;int i,j,x;scanf("%d",&t);while(t--){memset(p,0,sizeof(p));memset(a,0,sizeof(a));memset(b,0,sizeof(b));scanf("%d%d",&n,&m);for(i=0;i<n;i++){getchar();scanf("%d",&a[i]);}for(i=0;i<m;i++){getchar();scanf("%d",&b[i]);}x=Kmp();if(x<0) printf("-1\n");else printf("%d\n",x-m+1);}return 0;}