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  • HDU1711 ——Number Sequence(KMP模板题)

    Given two sequences of numbers : a[1], a[2], …… , a[N], and b[1], b[2], 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
    

    代码:

    #include <iostream>
    #include <cstdio>
    
    
    using namespace std;
    
    int next1[10005];
    int board[10005];
    int str[1000005];//寻找str串中是否有board串。
    int N,M;//分别是str和board中串的长度。
    
    void getNext(){
    
        int k = -1;
        int j = 0;
        next1[0] = -1;//next数组最好是从0开始要不然会很麻烦 
        while(j<M){
            if(k == -1 || board[j] == board[k]){
                if(board[j+1] == board[k+1]){ 
                    next1[++j] = next1[++k];
                }
                else next1[++j] = ++k;
            }
            else {
                k = next1[k];
            }
        }
    }
    
    
    
    int Find(){//返回匹配到的位置 
        getNext();//当str多次更改而board不变时可以拿出去,因为next数组只与所找的串有关。
        int temp1 = 0,temp2 = 0;
        while(temp1<N && temp2<M){
            if(temp2 == -1 || str[temp1] == board[temp2]){
                temp1++;temp2++;
            }
            else{
                temp2 = next1[temp2];
            }
        } 
        if(temp2 == M)return temp1-temp2;
        else return -1;  
    }
    
    int main(){
    
        int T;
        cin>>T;
        while(T--){
            scanf("%d %d",&N,&M);
            for(int i=0 ; i<N ; i++)scanf("%d",&str[i]); 
            for(int i=0 ; i<M ; i++)scanf("%d",&board[i]);//这里和next数组统一也从0开始 
            int re = Find();
            if(re == -1)printf("-1
    ");
            else printf("%d
    ",re+1);//由于前面是从0开始而题中是从1开始所以这里加1 
        }
    
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/vocaloid01/p/9514182.html
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