• hdu 1711 Number Sequence KMP


                                                                 Number Sequence

                                                        Time Limit:5000MS    Memory Limit:32768KB    64bit IO Format:%I64d & %I64u

    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  

    【一道没有不论什么杂质的纯KMP】

    #include <iostream>
    #include <cstdio>
    #include <cstring>
    using namespace std;
    const int N = 1000010;
    const int M = 10010;
    int nxt[M];
    int P[N], T[M];
    int n,m;
    void getnext(){
        int j, k;
        j = 0; k = -1; nxt[0] = -1;
        while(j<m){
            if (k==-1 || T[j]==T[k]){
                nxt[++j] = ++k;
            }
            else{
                k = nxt[k];
            }
        }
    }
    int kmp(){
        getnext();
        int j, k;
        j = 0; k = 0;
        while(k<m && j<n){
            if (k==-1 || T[k]==P[j]){
                ++k;++j;
    	    if(k==m) return j-m+1;
            }
            else{
                k = nxt[k];
            }
        }    
        return -1;
    }
    
    int main()
    {
        int t;
        scanf("%d", &t);
        while(t--){
            scanf("%d%d",&n,&m);
            for(int i=0;i<n;i++) scanf("%d",&P[i]);
            for(int i=0;i<m;i++) scanf("%d",&T[i]);
            printf("%d
    ", kmp());
        }
    
        return 0;
    }



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  • 原文地址:https://www.cnblogs.com/mthoutai/p/6834846.html
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