POJ3080

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Description

Input

Output

Sample Input

3
2
GATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATA
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
3
GATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATA
GATACTAGATACTAGATACTAGATACTAAAGGAAAGGGAAAAGGGGAAAAAGGGGGAAAA
GATACCAGATACCAGATACCAGATACCAAAGGAAAGGGAAAAGGGGAAAAAGGGGGAAAA
3
CATCATCATCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC
ACATCATCATAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
AACATCATCATTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTT

Sample Output

no significant commonalities
AGATAC
CATCATCAT

Source

仔细就好，了解了strstr函数。。。

/*
** 作者** 功能：查找最长公共字串
** poj3080
*/
#include<iostream>
#include<cstring>
using namespace std;
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        int n;
        cin>>n;
        int i,j,k;
        char str[12][65],ans[65],sub[65];
        for(i=0;i<n;i++)
            cin>>str[i];

        ans[0]='\0';
        for(i=3;i<=60;i++)//枚举字串长度
        {
            bool fd=false;//是否有公共字串判断
            for(j=0;j<=60-i;j++)//枚举字串起点
            {
                int len=0;//字串长度复制记录

                for(k=j;len<i;k++)
                {
                    sub[len++]=str[0][k];
                }
                sub[len]='\0';//取出字串gai
                bool ffdd=false;//该字串是否满足
                //接下来对比其他串
                for(k=1;k<n;k++)
                {
                    if(!strstr(str[k],sub))
                      {
                          ffdd=true;
                          break;
                      }
                }
                //如果该字串满足，则与ans比较并赋值
                if(!ffdd)
                {
                    if(strlen(ans)<strlen(sub))
                    {
                        strcpy(ans,sub);
                    }else if(strlen(ans)==strlen(sub)&&strcmp(sub,ans)<0)
                    {
                        strcpy(ans,sub);
                    }
                    fd=true;
                }
            }
            if(!fd)
                break;
        }
        if(strlen(ans)!=0)
        {
            cout<<ans<<endl;
        }else{
            cout<<"no significant commonalities"<<endl;
        }
    }
    return 0;
}

POJ3080

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原文地址：http://blog.csdn.net/fljssj/article/details/38925595