poj_3080

源链接：http://poj.org/problem?id=3080

题目大意就是找一个最长公共子串(长度要<3)，如果有多个，就输出字典序最小的那个。

因为数据较小，直接暴力即可。

暴力枚举某个字符串的长度为1的子串，长度为2的子串........，边枚举边记录符合条件的子串。

#include<stdio.h>
#include<iostream>
#include<string>
#include<string.h>
#include<algorithm>
#include<iomanip>
#include<vector>
#include<time.h>
#include<queue>
#include<stack>
#include<iterator>
#include<math.h>
#include<stdlib.h>
#include<limits.h>
#include<map>
//#define ONLINE_JUDGE
#define eps 1e-8
#define INF 0x7fffffff
#define FOR(i,a) for((i)=0;i<(a);(i)++)
#define MEM(a) (memset((a),0,sizeof(a)))
#define sfs(a) scanf("%s",a)
#define sf(a) scanf("%d",&a)
#define sfI(a) scanf("%I64d",&a)
#define pf(a) printf("%d\n",a)
#define pfI(a) printf("%I64d\n",a)
#define pfs(a) printf("%s\n",a)
#define sfd(a,b) scanf("%d%d",&a,&b)
#define sft(a,b,c)scanf("%d%d%d",&a,&b,&c)
#define for1(i,a,b) for(int i=(a);i<b;i++)
#define for2(i,a,b) for(int i=(a);i<=b;i++)
#define for3(i,a,b)for(int i=(b);i>=a;i--)
#define MEM1(a) memset(a,0,sizeof(a))
#define MEM2(a) memset(a,-1,sizeof(a))
const double PI=acos(-1.0);
template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
template<class T> inline T Min(T a,T b){return a<b?a:b;}
template<class T> inline T Max(T a,T b){return a>b?a:b;}
using namespace std;
#define ll __int64
int n,m;
#define Mod 1000000007
#define N 110
#define M 10010
char ch[15][65];
const int len = 60;
bool check(char *p){
	int len1 = strlen(p);
	int i;
	for(int s=1;s<n;s++){
		i=0;
		int flag = 0;
		while(i+len1<=len){
			int tmp = i;
			int j=0;
			while(p[j] == ch[s][tmp]&&j<len1){	//注意加上j<len1
				j++;
				tmp++;
			}
			if(j == len1){
				flag = 1;
				break;
			}
			i++;
		}
		if(!flag)
			return false;
	}
	return true;
}
int main(){
#ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
//  freopen("out.txt","w",stdout);
#endif
    	int t;
    	sf(t);
    	while(t--){
    		sf(n);
    		for(int i=0;i<n;i++)
    			sfs(ch[i]);
    		char tmp[65];
    		char ans[65];
    		memset(tmp,'\0',sizeof tmp);
    		memset(ans,'\0',sizeof ans);
    		for(int i=1;i<=len;i++){
    			for(int j=0;j<len;j++){
    				int s=0;
    				if(j+i>len) continue;
    				for(int k=j;k<j+i;k++)
    					tmp[s++] =ch[0][k];
    				tmp[s] = '\0';
    				if(check(tmp)){
    					int len1 = strlen(ans);
    					int len2 = strlen(tmp);
    					if(len1<len2)
    						memcpy(ans,tmp,sizeof(tmp));
    					else if(len1 == len2 && strcmp(tmp,ans)<0)
    						memcpy(ans,tmp,sizeof(tmp));
    				}
    			}
    		}
    		int ans_len = strlen(ans);
    		if(ans_len<3)
    			printf("no significant commonalities\n");
    		else
    			printf("%s\n",ans);
    	}
return 0;
}