题目:
输入样例:
3
2
GATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATA
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
3
GATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATA
GATACTAGATACTAGATACTAGATACTAAAGGAAAGGGAAAAGGGGAAAAAGGGGGAAAA
GATACCAGATACCAGATACCAGATACCAAAGGAAAGGGAAAAGGGGAAAAAGGGGGAAAA
3
CATCATCATCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC
ACATCATCATAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
AACATCATCATTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTT
输出样例:
no significant commonalities
AGATAC
CATCATCAT
大概题意:
输入n组,每组m个的字符串,找出m个字符串的最大公共字符串,如果这m个字符串的最大公共部分小于3输出no significant commonalities,如果大于或等于则输出该公共字符串,如果有多个相等的字符串就去找最小的。 |
Brute Force思路及算法:
#include<bits/stdc++.h>
using namespace std;
int main() {
int n,m;
char str[11][61],x[61],y[61];
int res_len,str_len,temp,i,j,k,t;
cin>>n;
while(n--) {
res_len=0;
cin>>m;
for(int i=0; i<m; i++) cin>>str[i];
str_len=strlen(str[0]);
for( i=1; i<=str_len; i++) {
for( j=0; j+i-1<str_len; j++) {
for( k=0; k<i; k++) {
x[k]=str[0][j+k];
}
x[k]='\0';
for( k=0; k<m; k++) {
if(strstr(str[k],x)==NULL) break;
}
if(k>=m) {
temp=strlen(x);
if(temp==res_len) {
for( t=0; t<res_len; t++) {
if(y[t]>x[t]) {
for( t=0; t<=res_len; t++) y[t]=x[t];
break;
}
}
}
if(temp>res_len) {
res_len=temp;
for( t=0; t<=res_len; t++) y[t]=x[t];
}
}
}
}
if(res_len>=3) cout<<y<<endl;
else cout<<"no significant commonalities"<<endl;
}
return 0;
}
特别提醒:
其实这个是因为,i在变,当i是1时,单个字符是要做60次遍历,但是当i大于等于2时,就是多个字符组成的字符串了,就做j+i-1次就行。就比如一个字符时在abcd这个字符串中遍历是a->b->c->d,要做四次。然而两个字符时就成了ab->bc->cd做三次遍历就够了。