HDU - 1711 - Number Sequence

先上题目：

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8442    Accepted Submission(s): 3858

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

 

　　题意就是给你两个数字序列，问你第二个序列在第一个序列的哪个位置第一次出现，如果没有出现过就输出-1。

　　这题其实就是一个kmp，字符改成了数字，而且要注意的是模式串有可能比原串长，这时候就直接输出-1。

 

上代码：

#include <stdio.h>
#include <string.h>
#define MAX 1000010
using namespace std;

int s[MAX],t[MAX],next[MAX],n,m;

void get_next()
{
    int i,k;
    memset(next,-1,sizeof(next));
    i=0;
    k=-1;
    while(i<m)
    {
        if(k==-1 || t[i]==t[k]){i++;k++;next[i]=k;}
        else k=next[k];
    }/*
    for(i=0;i<m;i++) printf("%d ",next[i]);
    printf("\n");
    */
}

int kmp()
{
    int i,j;
    i=0;j=0;
    while(i<n && j<m)
    {
        if(j==-1 || s[i]==t[j]){i++; j++;}
        else j=next[j];
    }
    if(j>=m) return i-m+1;
    else return -1;
}

int main()
{
    int T,i;
    //freopen("data.txt","r",stdin);
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d %d",&n,&m);
        for(i=0;i<n;i++) scanf("%d",&s[i]);
        for(i=0;i<m;i++) scanf("%d",&t[i]);
        if(m>n) printf("-1\n");
        else
        {
            get_next();
            int f=kmp();
            printf("%d\n",f);
        }
    }
    return 0;
}