poj3080 Blue Jeans

标签：kmp

Description

Input

Output

Sample Input

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

Sample Output

no significant commonalities
AGATAC
CATCATCAT
这题就是用暴力枚举，初看很难用枚举实现，其实复杂度不大的。
#include<stdio.h>
#include<string.h>
char s[13][70],s1[70],s2[70],ss[70];
int len[20],next[70],len1,len2;
void nextt()
{
	int i,j;
	i=0;j=-1;
	memset(next,-1,sizeof(next));
	while(i<len2){
		if(j==-1 || s2[i]==s2[j]){
			i++;j++;next[i]=j;
		}
		else j=next[j];
	}
}

int kmp()
{
	int i,j;
	i=0;j=0;
	while(i<len1 && j<len2){
		if(j==-1 || s1[i]==s2[j]){
			i++;j++;
		}
		else j=next[j];
	}
	if(j>=len2)return 1;
	else return 0;
}

int main()
{
	int n,m,i,j,T,cunzai,flag;
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d",&n);
		for(i=1;i<=n;i++){
			scanf("%s",s[i]);
			len[i]=strlen(s[i]);
		}
		cunzai=0;
		for(len2=60;len2>=3;len2--){
			memset(s2,0,sizeof(s2));
			for(j=0;j<=60-len2;j++){    //枚举起点 
				for(i=0;i<=len2-1;i++){
					s2[i]=s[1][j+i];
				}
				nextt();
				flag=1;
				for(i=2;i<=n;i++){
					len1=len[i];
					strcpy(s1,s[i]);
					if(!kmp()){
						flag=0;break;
					}
				}
				if(flag==1){
					if(cunzai==0){
						cunzai=1;strcpy(ss,s2);
					}
					else{
						if(strcmp(ss,s2)>0)strcpy(ss,s2);
					}
				}
			}
			if(cunzai)break;
		}
		if(cunzai)printf("%s\n",ss);
		else printf("no significant commonalities\n");
	}
	return 0;
}

poj3080 Blue Jeans

标签：kmp

原文地址：http://blog.csdn.net/kirito_acmer/article/details/46489931