poj 3080 Blue Jeans 暴力

Description

Input

Output

Sample Input

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

Sample Output

no significant commonalities
AGATAC
CATCATCAT

题意： 求出多组字符串的最大相邻公共子串，若长度小于三则输出“no significant commonalities”，否则，对于输出公共子串，对于同样长度的公共子串，输出按字母排序较小的串

思路：暴搜暴搜暴搜~~~

#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
char a[15][65];
char anss[65];
int main(){
    int n;
    scanf("%d",&n);
    while(n--){
        int m;
        scanf("%d",&m);
        int i,j,k;
        for(i=0;i<m;i++){
            scanf("%s",a[i]);
        }
        int ans=0;
        int len=strlen(a[0]);
        char sub[65];
        for(i=0;i<len;i++){
            for(j=i+2;j<len;j++){
                strncpy(sub,a[0]+i,j-i+1);
            sub[j-i+1]='\0';
            int flag=1;
            for(k=1;flag&&k<m;k++){
                if(strstr(a[k],sub)==0){
                    flag=0;
                }
            }
            if(flag&&(ans<strlen(sub)||(ans==strlen(sub)&&strcmp(anss,sub)>=0))){//记得比较一下
                ans=strlen(sub);
                strcpy(anss,sub);
            }
        }
        }
        if(ans<3) printf("no significant commonalities");
        else
        printf("%s",anss);
        printf("\n");
    }
}

poj 3080 Blue Jeans 暴力

原文：http://blog.csdn.net/taluoyaxin/article/details/47723779