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KMP算法 - 字符串匹配的简单运用 --- HDU 1711

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11606    Accepted Submission(s): 5294


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
 

 

Mean: 

 给你s1,s2两个串,让你找到s2在s1中出现的第一个位置。

analyse:

 KMP字符串水题,用c++写的,其实真正理解了KMP算法后,你会发现一点都不难。

Time complexity:O(n)  抓住:主串不回溯

 

Source code:

 

// Memory   Time// 1347K     0MS// by : Snarl_jsb// 2014-09-28-14.33#include<algorithm>#include<cstdio>#include<cstring>#include<cstdlib>#include<iostream>#include<vector>#include<queue>#include<stack>#include<map>#include<string>#include<climits>#include<cmath>#define N 1000010#define LL long longusing namespace std;int n,m;vector<int> a,b;vector<int> next;void GetNext(){    int k=0;    int len=b.size();    next.push_back(0);    for(int i=1;i<len;++i)    {        while(k!=0&&b[i]!=b[k])            k=next[k-1];        if(b[i]==b[k])            k++;        next.push_back(k);    }//    for(int i=0;i<next.size();++i)//        cout<<next[i]<<endl;}void KMP(){    GetNext();    bool flag=0;    int asize=a.size(),bsize=b.size(),k=0;    for(int i=0;i<asize;++i)    {        while(k>0&&a[i]!=b[k])        {            k=next[k-1];        }        if(a[i]==b[k])            k++;        if(k==m)        {            cout<<i-m+2<<endl;            flag=1;            break;        }    }    if(!flag) puts("-1");}int main(){    ios_base::sync_with_stdio(false);    cin.tie(0);//    freopen("C:\\Users\\ASUS\\Desktop\\cin.cpp","r",stdin);//    freopen("C:\\Users\\ASUS\\Desktop\\cout.cpp","w",stdout);    int t,tmp;    cin>>t;    while(t--)    {        cin>>n>>m;        a.clear(),b.clear(),next.clear();        for(int i=0;i<n;++i)        {            cin>>tmp;            a.push_back(tmp);        }        for(int i=0;i<m;++i)        {            cin>>tmp;            b.push_back(tmp);        }        KMP();    }    return 0;}

  

 

KMP算法 - 字符串匹配的简单运用 --- HDU 1711