hdu 1238 string handling violence

Substrings

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13779    Accepted Submission(s): 6689

http://acm.hdu.edu.cn/showproblem.php?pid=1238

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

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std;
const int maxn=100+3;
char str[maxn];
string ansstr,ss;
string str2[maxn];
int t,n;
int main(){
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        for(int i=0;i<n;i++){
            scanf("%s",str);
            str2[i]=str;
        }
        string temp=str2[0];
        int len=temp.size();
        int ans=0;
        for(int i=0;i<len;i++){
            for(int j=1;j+i<=len;j++){
                int cnt=1;
                ss=temp.substr(i,j);
                for(int k=1;k<n;k++){
                    if(str2[k].find(ss)!=-1){
                        cnt++;
                    }
                }
                if(cnt==n){
                    if(ans<j){
                        ans=j;
                        ansstr=ss;
                    }else if(ans==j){
                        ansstr=min(ansstr,ss);
                    }
                }
            }
        }
        reverse(temp.begin(),temp.end());
        for(int i=0;i<len;i++){
            for(int j=1;j+i<=len;j++){
                int cnt=1;
                ss=temp.substr(i,j);
                for(int k=1;k<n;k++){
                    if(str2[k].find(ss)!=-1){
                        cnt++; 
                    } 
                } 
                If (cnt == n) {
                     if (age < j) { 
                        years = j; 
                        ansstr = ss; 
                    } Else  if (years == j) { 
                        ansstr = min (ansstr, ss); 
                    } 
                } 
            } 
        } 
        Printf ( " % d \ n " , year); 
    } 
}

#include<stdio.h>//网上大佬的代码
#include<algorithm>
#include<iostream>
#include<string.h>
using namespace std;

const int maxn = 105;
char str[maxn];
string ansstr;
string str2[maxn];

int main()
{
    int n,t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(int i=0;i<n;i++)
        {
            scanf("%s",str);
            str2[i]=str;
        }
        int ans=0;
        string tmp=str2[0];
        int tmplen=tmp.size();
        for(int i=0;i<tmplen;i++)
        {
            for(int j=1;i+j<=tmplen;j++)
            {
                int cnt=1;
                string ss=tmp.substr(i,j);
                for(int k=1;k<n;k++)
                {
                    if(str2[k].find(ss)!=-1)
                        cnt++;
                }
                if(cnt==n)
                {
                    if(ans<j)
                    {
                        ans=j;
                        ansstr=ss;
                    }
                    else if(ans==j)
                    {
                        ansstr = min (ansstr, SS); 
                    } 
                } 
            } 
        } 
        Reverse (tmp.begin (), tmp.end ()); // Reverse () will be elements within the interval [beg, end) all reverse; 
        for ( int I = 0 ; I <the tmplen; I ++ ) 
        { 
            for ( int J = . 1 ; I + J <= the tmplen; J ++ ) 
            { 
                int CNT = . 1 ;
                 String SS = tmp.substr (I, J);
                 for ( int K = . 1 ; K <n-; K ++)
                {
                    if(str2[k].find(ss)!=-1)
                        cnt++;
                }
                if(cnt==n)
                {
                  //  cout<<ss<<" "<<cnt<<endl;
                    if(ans<j)
                    {
                        ans=j;
                        ansstr=ss;
                    }
                    else if(ans==j)
                    {
                        ansstr=min(ansstr,ss);
                    }
                }
            }
        }
        printf("%d\n",ans);
    }
}