题目链接：hihoCoder #1829 : Tomb Raider
时间限制:1000ms
单点时限:1000ms
内存限制:256MB
描述
Lara Croft, the fiercely independent daughter of a missing adventurer, must push herself beyond her limits when she discovers the island where her father disappeared. In this mysterious island, Lara finds a tomb with a very heavy door. To open the door, Lara must input the password at the stone keyboard on the door. But what is the password? After reading the research notes written in her father's notebook, Lara finds out that the key is on the statue beside the door.
The statue is wearing many arm rings on which some letters are carved. So there is a string on each ring. Because the letters are carved on a circle and the spaces between any adjacent letters are all equal, any letter can be the starting letter of the string. The longest common subsequence (let's call it "LCS") of the strings on all rings is the password. A subsequence is a sequence that can be derived from another sequence by deleting some or no elements without changing the order of the remaining elements.
For example, there are two strings on two arm rings: s1 = "abcdefg" and s2 = "zaxcdkgb". Then "acdg" is a LCS if you consider 'a' as the starting letter of s1, and consider 'z' or 'a' as the starting letter of s2. But if you consider 'd' as the starting letter of s1 and s2, you can get "dgac" as a LCS. If there are more than one LCS, the password is the one which is the smallest in lexicographical order.
Please find the password for Lara.
输入
There are no more than 10 test cases.
In each case:
The first line is an integer n, meaning there are n (0 < n ≤ 10) arm rings.
Then n lines follow. Each line is a string on an arm ring consisting of only lowercase letters. The length of the string is no more than 8.
输出
For each case, print the password. If there is no LCS, print 0 instead.
样例输入
2 abcdefg zaxcdkgb 5 abcdef kedajceu adbac abcdef abcdafc 2 abc def
样例输出
acdg acd 0
题意：给n个字符串，首尾相连，求最长公共子序列，若答案多个，输出字典序最小的
思路：暴力把每个字符串的每个子序列放入map（不可重复放入），出现n次的就是公共的子序列，符合条件的公共子序列都放入数组，最后排序，得到最长中字典序最小的那个
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<iostream>
#include<map>
#include<vector>
#include<string>
using namespace std;
string s;
string newstr(int pos){
string news1(s,pos);
string news2(s,0,pos);
string news=news1+news2;
return news;
}
bool cmp(string a,string b){
if(a.length()!=b.length())return a.length()>b.length();
return a<b;
}
int main(){
int n;
while(~scanf("%d",&n)){
map<string,int>mp,vis;
map<string,int>::iterator it;
for(int i=1;i<=n;i++){
cin>>s;
vis.clear();
int len=s.length();
for(int j=0;j<len;j++){//遍历任意位置开头的字符串
string news=newstr(j);
for(int k=1;k<(1<<len);k++){//遍历一个字符串的每个子序列
string str;
for(int p=0;p<len;p++){
if(k>>p&1){
str+=news[p];
}
}
if(vis[str]==0){//同一个字符串，在遍历中会有重复子序列出现，要排除
vis[str]=1;
mp[str]++;
}
}
}
}
int flag=0;
vector<string>ans;
for(it=mp.begin();it!=mp.end();it++){
if(it->second==n){//出现n次，符合条件，放入ans
ans.push_back(it->first);
flag=1;
}
}
sort(ans.begin(),ans.end(),cmp);
if(!flag)cout<<0<<endl;
else cout<<ans[0]<<endl;
}
return 0;
}