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  • HDU1711 KMP(模板题)

    Number Sequence

    Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 24587    Accepted Submission(s): 10436


    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

     题意:

    给出T和P求p出现的位置

    代码:

    #include<iostream>
    #include<cstdio>
    #include<cstring>
    using namespace std;
    const int maxn=1000006;
    const int maxm=10004;
    int p[maxm],t[maxn];
    int f[maxm],n,m;
    void getfail(int* p,int* f){
        f[0]=0;f[1]=0;
        for(int i=1;i<m;i++){
            int j=f[i];
            while(j&&p[i]!=p[j]) j=f[j];
            f[i+1]=(p[i]==p[j]?j+1:0);
        }
    }
    int find(int* t,int* p,int* f){
        getfail(p,f);
        int j=0;
        for(int i=0;i<n;i++){
            while(j&&p[j]!=t[i]) j=f[j];
            if(p[j]==t[i]) j++;
            if(j==m) return i-m+1+1;//下标从0开始
        }
        return -1;
    }
    int main()
    {
        int cas;
        scanf("%d",&cas);
        while(cas--){
            scanf("%d%d",&n,&m);
            for(int i=0;i<n;i++) scanf("%d",&t[i]);
            for(int i=0;i<m;i++) scanf("%d",&p[i]);
            int ans=find(t,p,f);
            printf("%d
    ",ans);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/--ZHIYUAN/p/6411227.html
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