poj 2337-Catenyms

Catenyms

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 9832 Accepted: 2561

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:

dog.gophergopher.ratrat.tigeraloha.alohaarachnid.dog

A 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

26alohaarachniddoggopherrattiger3oakmapleelm

Sample Output

aloha.arachnid.dog.gopher.rat.tiger***代码：
#include <iostream>#include <cmath>#include <cstdio>#include <cstring>using namespace std;int degreein[26],degreeout[26],d[1005];bool visit[1005];struct deg{int start;int end;char w[30];}deg_order[1005];int num;int cmp(const void *a,const void *b){deg *c=(deg *)a;deg *d=(deg *)b;if(c->start!=d->start)return c->start-d->start;elsereturn strcmp(c->w,d->w);}int Abs(int a){if(a>0)return a;elsereturn -a;}int getsource()//寻找起点{int i;int source,cnt1=0,cnt2=0;for(i=0;i<26;i++){if(Abs(degreein[i]-degreeout[i])==2)return -1;else if(degreein[i]-degreeout[i]==1){cnt1++;//基度顶点加1}else if(degreein[i]-degreeout[i]==-1){cnt2++;//基度顶点加1source=i;}}if(cnt1>1||cnt2>1)//如果基度顶点数大于一说明不能构成欧拉路径return -1;if(cnt1==0)//如果不存在基度顶点{int i;for(i=0;i<26;i++)if(degreeout[i])return i;}else     //找到起始点return source;}bool dfs(int source,int cnt)//搜索是否存在欧拉路径{int i;if(cnt==num)return true;for(i=0;i<num;i++){if(visit[i]||deg_order[i].start<source)continue;elseif(deg_order[i].start>source)return false;d[cnt]=i;visit[i]=1;if(dfs(deg_order[i].end,cnt+1))return true;visit[i]=0;}return false;}int main(){int t;cin>>t;while(t--){memset(visit,0,sizeof(visit));memset(degreein,0,sizeof(degreein));memset(degreeout,0,sizeof(degreeout));cin>>num;int i;for(i=0;i<num;i++){cin>>deg_order[i].w;deg_order[i].start=deg_order[i].w[0]-'a';deg_order[i].end=deg_order[i].w[strlen(deg_order[i].w)-1]-'a';degreein[deg_order[i].end]++;degreeout[deg_order[i].start]++;}int source=getsource();if(source==-1){cout<<"***"<<endl;continue;}qsort(deg_order,num,sizeof(deg_order[0]),cmp);if(!dfs(source,0)){            cout<<"***"<<endl;continue;}int j;cout<<deg_order[d[0]].w;for(j=1;j<num;j++)cout<<'.'<<deg_order[d[j]].w;cout<<endl;}return 0;}