Catenyms
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 8368 | Accepted: 2202 |
Description
A catenym is a pair of words separated by a period such that the last letter of the first word is the same as the last letter of the second. For example, the following are catenyms:
aloha.aloha.arachnid.dog.gopher.rat.tiger
Given a dictionary of lower case words, you are to find a compound catenym that contains each of the words exactly once.
dog.gopher gopher.rat rat.tiger aloha.aloha arachnid.dogA compound catenym is a sequence of three or more words separated by periods such that each adjacent pair of words forms a catenym. For example,
aloha.aloha.arachnid.dog.gopher.rat.tiger
Given a dictionary of lower case words, you are to find a compound catenym that contains each of the words exactly once.
Input
The first line of standard input contains t, the number of test cases. Each test case begins with 3 <= n <= 1000 - the number of words in the dictionary. n distinct dictionary words follow; each word is a string of between 1 and 20 lowercase letters on a line by itself.
Output
For each test case, output a line giving the lexicographically least compound catenym that contains each dictionary word exactly once. Output "***" if there is no solution.
Sample Input
2 6 aloha arachnid dog gopher rat tiger 3 oak maple elm
Sample Output
aloha.arachnid.dog.gopher.rat.tiger ***
用DFS实现的
1 #include <iostream> 2 #include <stdio.h> 3 #include <algorithm> 4 #include <string.h> 5 #include <math.h> 6 #include <vector> 7 #include <stack> 8 #include <map> 9 using namespace std; 10 #define ll long long int 11 #define INF 5100000 12 string a[1500]; 13 int v[2000]; 14 vector<pair<int,int> >aa[30]; 15 vector<int>ab; 16 int p[30]; 17 int d1[30]; 18 int d2[30]; 19 int find(int x) 20 { 21 while(x!=p[x]) 22 x=p[x]; 23 return x; 24 } 25 void fun(int x,int eid) 26 { 27 for(int i=0; i<aa[x].size(); i++) 28 if(!v[aa[x][i].second]) 29 { 30 v[aa[x][i].second]=true; 31 fun(aa[x][i].first,aa[x][i].second); 32 } 33 if(eid>0)ab.push_back(eid); 34 } 35 int main() 36 { 37 int tt; 38 int i,j; 39 cin>>tt; 40 for(i=0; i<tt; i++) 41 { 42 int n; 43 cin>>n; 44 memset(d1,0,sizeof(d1)); 45 memset(d2,0,sizeof(d2)); 46 for(j=1; j<30; j++) 47 p[j]=j,aa[j].clear(); 48 for(j=1; j<=n; j++) 49 { 50 cin>>a[j]; 51 } 52 sort(a+1,a+n+1); 53 for(j=1; j<=n; j++) 54 { 55 int x=a[j][0]-'a'+1; 56 int y=a[j][a[j].length()-1]-'a'+1; 57 d1[x]++; 58 d2[y]++; 59 aa[x].push_back(make_pair(y,j)); 60 int fx=find(x); 61 int fy=find(y); 62 if(fy!=fx) 63 p[fy]=fx; 64 } 65 int sum=0; 66 int start=a[1][0]-'a'+1; 67 for(j=1; j<30; j++) 68 { 69 if(d1[j]==0&&d2[j]==0)continue; 70 if(p[j]==j)sum++; 71 } 72 if(sum>1) 73 cout<<"***"<<endl; 74 else 75 { 76 int dd1=0,dd2=0; 77 for(j=1; j<30; j++) 78 { 79 if(d1[j]!=d2[j]) 80 { 81 if(d1[j]-d2[j]==1) 82 dd1++,start=j; 83 else if(d2[j]-d1[j]==1) 84 dd2++; 85 else 86 { 87 dd2=199; 88 break; 89 } 90 } 91 } 92 if(dd2<=1&&dd1<=1) 93 { 94 memset(v,0,sizeof(v)); 95 ab.clear(); 96 fun(start,0); 97 for( j=ab.size()-1; j>=0; j--) 98 { 99 cout<<a[ab[j]]; 100 if(j!=0)cout<<"."; 101 else cout<<" "; 102 } 103 104 } 105 else cout<<"***"<<endl; 106 } 107 } 108 }