Q - Number Sequence

淩亂°似流年 2022-03-21 13:20 176阅读 0赞

### 题目描述： ###

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.  
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：  
给出两个整形序列，判断第二个整形序列是不是第一个整形序列的子序列，如果是，那么请输出是从哪一个位置开始的子序列。

##### 分析: #####

本题就是一个简单字符串匹配问题，不过将字符串改成了整形数。至于怎么输出是从哪一个位置开始的，其实很简单，我们在匹配成功时的i值是子序列的结束位置，那么子序列的长度又是可知的，直接用结束位置减去长度就是开始位置。

#include"stdio.h"
    #include"string.h"
    void Get_next(int text[],int next[],int m)
    {
        int i,j;
        int len;
        next[0]=-1;
        i=0;
        j=-1;
        while(i<m)
        {
            if(j==-1||text[i]==text[j])
            {
                i++;
                j++;
                if(text[i]==text[j])
                    next[i]=next[j];
                else
                    next[i]=j;
            }
            else
                j=next[j];
        }
    }
    int Index_KMP(int text[],int next[],int word[],int n,int m)
    {
        int i,j,k;
        int lentext=n;
        int lenword=m;
        int count=1;
        i=0;
        j=0;
            while(j<lenword&&i<lentext)
            {
                if(j==-1||text[i]==word[j])
                {
                    i++;
                    j++;
                }
    
                else
                {
                    j=next[j];
                }
    
    
            }
      if(j==lenword)
         return i-lenword+1;
        return -1;
    }
        int text[1000001],word[1000001];
        int next[1000001];
    int main()
    {
    
        int i,j,count,T,m,n;
        while(~scanf("%d",&T))
        {
            while(T--)
            {
                scanf("%d%d",&n,&m);
                for(i=0;i<n;i++)
                    scanf("%d",&text[i]);
                for(i=0;i<m;i++)
                    scanf("%d",&word[i]);
                Get_next(word,next,m);
                count=Index_KMP(text,next,word,n,m);
                    printf("%d\n",count);
    
            }
        }
    }