HDU1711 Number Sequence

来源：互联网发布：查看本机mac地址编辑：程序博客网时间：2024/06/06 17:14

Number Sequence

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 24798 Accepted Submission(s): 10499

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

由于这道题的测试数据规模较大，传统的模式匹配算法一定会超限，所以我们可以使用KMP算法来实现这个功能。

KMP算法虽然算法简单，但是并不好理解，所以我就找了一个相关的链接可供参考。

点击打开链接

#include<stdio.h>#include<string.h>#define N 1000005int next[N],a[N],b[N];int m,n;void Next()//就是上面的分匹配表的实现{    int i,j;    i=0;    j=-1;    next[i]=j;   //匹配表初值    while(i<m)    {        if(j==-1||b[i]==b[j])        {            i++;            j++;            next[i]=j;        }        else            j=next[j];    }    return ;}int KMP()//kmp匹配算法{    int i,j;    i=j=0;    Next();//先计算部分匹配表    while(i<n)    {        if(j==-1||a[i]==b[j])        {            i++;            j++;            if(j==m)                return i-m+1;//找到目标字符串，返回到主程序。        }        else            j=next[j];//a[i]与b[j]不匹配，查表需要跳过的字符个数。    }    return -1;//没有找到返回-1}int main(){    int t;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&n,&m);        for(int i=0; i<n; i++)            scanf("%d",&a[i]);        for(int i=0; i<m; i++)            scanf("%d",&b[i]);        if(KMP()==-1)            printf("-1\n");        else            printf("%d\n",KMP());    }    return 0;}

0 0