#1829 : Tomb Raider

時間限制:1000ms

單點時限:1000ms

內存限制: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.

輸入

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.

輸出

For each case, print the password. If there is no LCS, print 0 instead.

樣例輸入

2 abcdefg zaxcdkgb 5 abcdef kedajceu adbac abcdef abcdafc 2 abc def

樣例輸出

acdg acd 0 #include<bits/stdc++.h> using namespace std;#define e exp(1) #define pi acos(-1) #define mod 998244353 #define inf 0x3f3f3f3f #define ll long long #define ull unsigned long long #define mem(a,b) memset(a,b,sizeof(a)) int gcd(int a,int b){return b?gcd(b,a%b):a;}int n,flag; string s; map<string,int> m,vis; vector<string> v; string sub(int x) {string index=s.substr(x);index+=s.substr(0,x);return index; } bool cmp(string a,string b) {if(a.length()==b.length())return a<b;return a.length()>b.length(); } int main() {while(~scanf("%d",&n)){m.clear();v.clear();flag=0;for(int cas=0; cas<n; cas++){cin>>s;int len=s.length();for(int l=0; l<len; l++){string s1=sub(l);for(int i=1; i<(1<<len); i++){string s2;for(int j=0; j<len; j++){if((1<<j)&i)s2+=s1[j];}if(vis[s2]==0){m[s2]++;vis[s2]=1;}}}vis.clear();}for(map<string,int>::iterator i=m.begin(); i!=m.end(); i++){if(i->second==n){v.push_back(i->first);flag=1;}}sort(v.begin(), v.end(), cmp);if(flag==0)puts("0");else cout<<v[0]<<endl;}return 0; }

?

總結

以上是生活随笔為你收集整理的ACM/ICPC 2018亚洲区预选赛北京赛站网络赛 Tomb Raider（map+二进制枚举）的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。