hdu 1238 Substrings KMP问题

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给出一些字符串，找出最大满足条件的连续子串使之满足：在每一个字符串中都正序存在或逆序存在。输出最大长度。

Substrings

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8758 Accepted Submission(s): 4088

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

23ABCDBCDFFBRCD2roseorchid

Sample Output

Author

Asia 2002, Tehran (Iran), Preliminary

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我的做法是把第一个字符串当作缝衣针，其他字符串先翻转，与自身组成两人一组。

对于每一组求出缝衣针中每个位置能匹配的最大长度。

之后求出所有非缝衣针字符集合中缝衣针字符串每个位置能匹配的最大长度。

取所有位置的最大值。

/**========================================== *   This is a solution for ACM/ICPC problem * *   @source：hdu 1238 Substrings *   @type:  data_structrue KMP *   @author: wust_ysk *   @blog:  http://blog.csdn.net/yskyskyer123 *   @email: 2530094312@qq.com *===========================================*/#include<cstdio>#include<string>#include<cstring>#include<iostream>#include<cmath>#include<algorithm>using namespace std;typedef long long ll;const int INF =0x3f3f3f3f;const int maxn=100    ;//const int maxV=12    ;string N[2*maxn+5];string M;string S;int  nex[maxn+5];int len[2*maxn+5];int len_S;int len_M,n;int dp[2*maxn][maxn+5];void get(int ind){    int p=ind-1;    N[ind]="";    for(int i=len[p]-1;i>=0;i--)    {        N[ind]+=N[p][i];    }}void getnex(){    nex[0]=nex[1]=0;    for(int i=1;i<len_M;i++)    {        int p=nex[i];        while(p&&M[p]!=M[i])  p=nex[p];        nex[i+1]= M[p]==M[i]? p+1:0;    }}void KMP(int ind,int delta){    int p=0;    for(int i=0;i<len[ind];i++)    {        while(p&&M[p]!=N[ind][i])  p=nex[p];        if(M[p]==N[ind][i])        {            dp[ind][delta+p]=max(dp[ind][delta+p],p+1 );            p++;        }        if(p==len_M)  p=nex[p];    }}void getmax(int ind){    for(int i=0;i<len_S;i++)    {        dp[ind][i]=max(dp[ind][i],dp[ind+1][i]);    }}void work(){    memset(dp,0,sizeof dp);     for(int st=0;st<len_S;st++)    {        M=S.substr(st,len_S);        len_M=len_S-st;        getnex();        for(int i=0;i<n;i++)      {        KMP( 2*i+1,st );        KMP( 2*i+2,st );      }  }  for(int i=0;i<n;i++)  {      getmax(2*i+1);  }  int ans=0;  memset(dp[0],0x3f,sizeof dp[0]);    for(int j=0;j<len_S;j++)    {        for(int i=1;i<=2*n-1;i+=2)        {             dp[0][j]=min(dp[0][j],dp[i][j]  );        }        ans=max(ans,dp[0][j]);    }    printf("%d\n",ans);}int main(){    int T;cin>>T;    while(T--)    {        cin>>n;        cin>>S;        len_S=S.length();        if(--n==0)  {printf("%d\n",len_S);continue;}        for(int i=0;i<n;i++)        {            cin>>N[2*i+1];            len[2*i+2]=len[2*i+1]=N[2*i+1].length();            get(2*i+2);        }        work();    }   return 0;}

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