Catenyms
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 11648 | Accepted: 3036 |
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.gopher gopher.rat rat.tiger aloha.aloha arachnid.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
2 6 aloha arachnid dog gopher rat tiger 3 oak maple elm
Sample Output
aloha.arachnid.dog.gopher.rat.tiger *** 题意:给定一系列字符串,如果某个字符串的结尾和另外一个字符串的开头相等,那么两个字符串就可以拼接在一起,现在问所有的字符串能否刚好都出现一次?如果存在,输出字典序最小的那个组合. 题解:将字符串看成一条边,然后将其起始字符和结尾字符看成边的两个端点,构造一个有向图,然后就判断这个图是否为欧拉图了.判断单向欧拉图的方法:
有向欧拉通路：起点：出度-入度=1，终点：入度-出度=1，其它点：入度==出度
有向欧拉回路：所有点：入度==出度
还要利用并查集判断一下这个图是否只有一个连通分量.
然后将所有的边按照字典序排序,找到起点,进行DFS(Edge &e = edge[u][i] 这里检查了半天,一点要记得是地址啊...),即可得到答案.
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <math.h>
#include <vector>
#include <stack>
using namespace std;
const int N = 2222;
char str[100];
struct Edge{
char str[100];
int to,del;
};
typedef vector <Edge> vec;
stack <string> ans;
vec edge[N];
int in[N],out[N],father[N];
bool vis[N],mark[30];
int n;
void init(){
memset(in,0,sizeof(in));
memset(out,0,sizeof(out));
memset(vis,false,sizeof(vis));
for(int i=1;i<=30;i++){
mark[i] = false;
edge[i].clear();
father[i] = i;
}
}
int _find(int x){
return x==father[x]?x:father[x] = _find(father[x]);
}
void dfs(int u){
for(int i=0;i<edge[u].size();i++){
Edge &e = edge[u][i]; ///这里要取地址
int v = e.to;
if(!e.del){
e.del = 1;
dfs(v);
ans.push(e.str);
}
}
}
bool cmp(Edge a,Edge b){
return strcmp(a.str,b.str)<0;
}
int main()
{
int tcase;
scanf("%d",&tcase);
while(tcase--){
init();
scanf("%d",&n);
int S = N;
for(int i=1;i<=n;i++){
scanf("%s",str);
int s = str[0]-‘a‘+1;
int t = str[strlen(str)-1]-‘a‘+1;
Edge e;
e.del = 0;
strcpy(e.str,str);
e.to = t;
edge[s].push_back(e);
in[t]++;
out[s]++;
mark[t] = mark[s] = 1;
int u = _find(s);
int v = _find(t);
if(u!=v) father[u] = v;
S = min(S,min(s,t));
}
int num1 = 0,num2 = 0,num3 = 0,num4 = 0;
for(int i=1;i<=26;i++){
if(mark[i]==1&&father[i]==i) num4++;
if(out[i]-in[i]==1){
S = i;
num1++;
}
else if(in[i]-out[i]==1){
num2++;
}else if(in[i]-out[i]!=0){
num3++;
}
if(!edge[i].empty())
sort(edge[i].begin(),edge[i].end(),cmp);
}
if(num3||!(num1==1&&num2==1)&&!(num1==0&&num2==0)||num4>1){
printf("***\n");
continue;
}
dfs(S);
int flag = 0;
while(!ans.empty()){
if(flag) printf(".");
cout<<ans.top();
flag = 1;
ans.pop();
}
printf("\n");
}
}