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  • Substrings kmp

      

    Problem Description
    You are given a number of case-sensitive strings of alphabetic characters, find the largest string X, such that either X, or its inverse can be found as a substring of any of the given strings.
     
    Input
    The first line of the input file contains a single integer t (1 <= t <= 10), the number of test cases, followed by the input data for each test case. The first line of each test case contains a single integer n (1 <= n <= 100), the number of given strings, followed by n lines, each representing one string of minimum length 1 and maximum length 100. There is no extra white space before and after a string.
     
    Output
    There should be one line per test case containing the length of the largest string found.
     
    Sample Input
    2 3 ABCD BCDFF BRCD 2 rose orchid
     
    Sample Output
    2 2
     
    直接暴力kmp即可  
    有一个函数 :reverse(p2.begin(),p2.end()); 可以直接反转 
    注意如果没有则输出0;
     
    第一次的写法  虽然艰难的过了 但是时间为500ms
     
    #include<bits/stdc++.h>
    using namespace std;
    //input
    #define rep(i,a,b) for(int i=(a);i<=(b);i++)
    #define repp(i,a,b) for(int i=(a);i>=(b);i--)
    #define RI(n) scanf("%d",&(n))
    #define RII(n,m) scanf("%d%d",&n,&m);
    #define RIII(n,m,k) scanf("%d%d%d",&n,&m,&k)
    #define RS(s) scanf("%s",s);
    #define ll long long
    #define inf 0x3f3f3f3f
    #define REP(i,N)  for(int i=0;i<(N);i++)
    #define CLR(A,v)  memset(A,v,sizeof A)
    //////////////////////////////////
    #define N 100+5
    #define mod 10007
    string s,p;
    int can[105];
    string temp[N];
    int nex[N];
    string str;
    int lenp,lens;
    void getnext()
    {
        nex[0]=-1;
        int k=-1,j=0;
        while(j<lenp-1)
        {
            if(k==-1||p[k]==p[j])
                nex[++j]=++k;
            else k=nex[k];
        }
    }
    int kmp(string s)
    {
      int lens=s.size();
      int j=0,i=0;
      while(i<lens&&j<lenp)
      {
          if(s[i]==p[j]||j==-1)
          {
              i++;
              j++;
          }
          else j=nex[j];
          if(j==lenp)
          {
            return 1;
          }
      }
    return 0;
    }
    int main()
    {
    
        int cas;
        RI(cas);
        while(cas--)
        {
            int n;RI(n);
            int minn=inf;
            rep(i,1,n)
            {
            cin>>temp[i];
            if(temp[i].size()<minn )minn=temp[i].size();
            }
            rep(i,1,n)
            if(temp[i].size()==minn)
            {
                str=temp[i];
                break;
            }
            int end1=0;
            int maxx=0;
            for(lenp=minn;lenp>=1;lenp--)
            {
                if(end1)break;
                for(int j=0;j<=minn;j++)
                if(j+lenp-1<minn)
                {
                    int ok=1;
                    p=str.substr(j,lenp);
                    getnext();
                    rep(i,1,n)
                    {
                        can[i]=0;
    
                        if(kmp(temp[i]))
                            can[i]=1;
                    }
                    reverse(p.begin(),p.end());
                    getnext();
                    rep(i,1,n)
                    {
    
                        if(kmp(temp[i]))
                            can[i]=1;
    
                        if(can[i]==0)
                        {
                            ok=0;break;
                        }
                    }
                    if(ok)
                    {
                        maxx=lenp;end1=1;
                        break;
                    }
                }
            }
            cout<<maxx<<endl;
        }
        return 0;
    }
    View Code
     
    下面的为70ms
     因为其实大部分情况下是不匹配的  匹配的情况微乎其微   所以直接两个一起判断即可  即使每次判断都要更行next
     
    #include<bits/stdc++.h>
    using namespace std;
    //input
    #define rep(i,a,b) for(int i=(a);i<=(b);i++)
    #define repp(i,a,b) for(int i=(a);i>=(b);i--)
    #define RI(n) scanf("%d",&(n))
    #define RII(n,m) scanf("%d%d",&n,&m);
    #define RIII(n,m,k) scanf("%d%d%d",&n,&m,&k)
    #define RS(s) scanf("%s",s);
    #define ll long long
    #define inf 0x3f3f3f3f
    #define REP(i,N)  for(int i=0;i<(N);i++)
    #define CLR(A,v)  memset(A,v,sizeof A)
    //////////////////////////////////
    #define N 100+5
    #define mod 10007
    string s,p;
    int can[105];
    string temp[N];
    int nex[N];
    string str;
    int lenp,lens;
    void getnext()
    {
        nex[0]=-1;
        int k=-1,j=0;
        while(j<lenp-1)
        {
            if(k==-1||p[k]==p[j])
                nex[++j]=++k;
            else k=nex[k];
        }
    }
    int kmp(string s)
    {
      int lens=s.size();
      int j=0,i=0;
      while(i<lens&&j<lenp)
      {
          if(s[i]==p[j]||j==-1)
          {
              i++;
              j++;
          }
          else j=nex[j];
          if(j==lenp)
          {
            return 1;
          }
      }
    return 0;
    }
    int main()
    {
        int cas;
        RI(cas);
        while(cas--)
        {
            int n;RI(n);
            int minn=inf;
            rep(i,1,n)
            {
            cin>>temp[i];
            if(temp[i].size()<minn )minn=temp[i].size();
            }
            rep(i,1,n)
            if(temp[i].size()==minn)
            {
                str=temp[i];
                break;
            }
            int maxx=0;
            int end1=0;
            for(lenp=minn;lenp>=1;lenp--)
            {
                if(end1)break;
                for(int j=0;j<=minn;j++)
                if(j+lenp-1<minn)
                {
                    string p1=str.substr(j,lenp);
                    string p2=p1;
                    reverse(p2.begin(),p2.end());
    
                    int i;
                    for(i=1;i<=n;i++)
                    {
                        int flag=0;
                        p=p1;
                        getnext();
                        if(kmp(temp[i]))
                            flag=1;
                        p=p2;
                        getnext();
                        if(kmp(temp[i]))
                            flag=1;
                        if(flag==0)
                        break;
                    }
                    if(i==n+1&&maxx<lenp)
                        maxx=lenp,end1=1;
                }
            }
            cout<<maxx<<endl;
        }
        return 0;
    }
    View Code
     
     
     
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  • 原文地址:https://www.cnblogs.com/bxd123/p/10673749.html
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