hdu 1711 Number Sequence KMP（水）

AI音视频处理技术概览

最新推荐文章于 2019-07-14 13:24:47 发布

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最新推荐文章于 2019-07-14 13:24:47 发布

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本文概述了AI音视频处理领域的关键技术，包括视频分割、语义识别、自动驾驶、AR、SLAM等，并探讨了其在实际应用中的作用。

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1711

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

模板~~

代码：

#include <algorithm>
#include <cstdlib>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cctype>
#include <cmath>
#include <stack>
#include <queue>
#include <list>
#include <map>
#include <set>

using namespace std;

#define min2(x, y)     min(x, y)
#define max2(x, y)     max(x, y)
#define min3(x, y, z)  min(x, min(y, z))
#define max3(x, y, z)  max(x, max(y, z))
#define clr(x, y)      memset(x, y, sizeof(x))
#define fr(i,n)        for(int i = 0; i < n; i++)
#define fr1(i,n)       for(int i = 1; i < n; i++)
#define upfr(i,j,n)    for(int i = j; i <= n; i++)
#define dowfr(i,j,n)   for(int i = n; i >= j; i--)
#define scf(n)         scanf("%d", &n)
#define scf2(n,m)      scanf("%d %d",&n,&m)
#define scf3(n,m,p)    scanf("%d %d %d",&n,&m,&p)
#define ptf(n)         printf("%d",n)
#define ptf64(n)       printf("%I64d",n)
#define ptfs(s)        printf("%s",s)
#define ptln()         printf("\n")
#define ptk()          printf(" ")
#define ptc(c)         printf("%c",c)
#define srt(a,n)       sort(a,n)
#define LL long long
#define pi acos(-1.0)
#define eps 0.00001
#define maxn 1000005
#define mod 10000007

int str1[maxn], str2[maxn];
int nex[maxn];
int n,m;

void get_next(int str[])
{
    int i = 1, j = 0;
    nex[0] = -1;
    while(i < m)
    {
        if(j == -1 || str[i] == str[j])
        {
            i ++;
            j ++;
            nex[i] = j;
        }
        else
            j = nex[j];
    }
}

int kmp(int str1[], int str2[])
{
    int i = 0, j = 0;
    get_next(str2);
    while(i < n && j < m)
    {
        if(j == -1 || str1[i] == str2[j])
        {
            i ++;
            j ++;
        }
        else
            j = nex[j];
    }
    if(j == m)
        return i - j;
    return -1;
}
int main()
{
    //freopen("in.txt","r",stdin);
    int ca;
    scf(ca);
    while(ca--)
    {
        clr(nex, -1);
        clr(str1, 0);
        clr(str2, 0);
        scf2(n, m);
        int cnt1 = 0, cnt2 = 0;
        int k;
        fr(i, n)
        {
            scf(str1[i]);
        }
        fr(i, m)
        {
           scf(str2[i]);
        }
        int res = kmp(str1, str2);
        if(res != -1)
            printf("%d\n", res + 1);
        else
            printf("-1\n");
    }
    return 0;