POJ-2337 Catenyms

Catenyms

Time Limit: 1000MS Memory Limit: 65536K   

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***

————————————————————独立的分割线————————————————————

前言：还是欧拉路，只不过不像是旋转鼓轮那样保证有欧拉路了，这次要先判断。

思路：边权就是该单词，（因此写得很矬），首字母和尾字母就是点。数入度出度、判欧拉，显式栈DFS、输出路径。题目要求字典序，这个好办，先把单词全部读进来，然后存进结构体快排。再建图即可。

代码如下：

/*ID: j.sure.1PROG:LANG: C++*//****************************************/#include <cstdio>#include <cstdlib>#include <cstring>#include <algorithm>#include <cmath>#include <stack>#include <queue>#include <vector>#include <map>#include <string>#include <climits>#include <iostream>#define INF 0x3f3f3f3fusing namespace std;/****************************************/const int M = 1e3+5, N = 26;int tot, head[N], pi, id[N], od[N], m, st;char path[M][25];struct Node {int v, next;char w[25];}edge[M];struct STA {int u;char w[25];}sta[M];struct Str {char str[25];}s[M];bool vis[N];int cmp(Str a, Str b){return strcmp(a.str, b.str) >= 0;}bool judge(){st = -1;int one = 0;for(int i = 0; i < 26; i++) if(vis[i]) {if(st == -1) st = i;if(abs(od[i]-id[i]) >= 2) return false;if(abs(od[i]-id[i]) == 1) {one++;if(od[i] > id[i]) st = i;if(one > 2) return false;}}if(one == 1) return false;return true;}void add(int u, int v, char *w){edge[tot].v = v;strcpy(edge[tot].w, w);edge[tot].next = head[u];head[u] = tot++;}void Euler(){int top = -1;sta[0].u = st; sta[0].w[0] = '#';top++;while(top != -1) {int u = sta[top].u, i;char w[25]; strcpy(w, sta[top].w);for(i = head[u]; i != -1; i = edge[i].next) {if(edge[i].v != -1) {int v = edge[i].v;edge[i].v = -1;top++;sta[top].u = v;strcpy(sta[top].w, edge[i].w);break;}}if(i == -1) {if(w[0] != '#') strcpy(path[pi++], w);top--;}}}int main(){#ifdef J_Sure//freopen("000.in", "r", stdin);//freopen(".out", "w", stdout);#endifint T;scanf("%d", &T);while(T--) {scanf("%d", &m);tot = 0;memset(head, -1, sizeof(head));memset(id, 0, sizeof(id));memset(od, 0, sizeof(od));memset(vis, 0, sizeof(vis));char w[25];int u, v;for(int i = 0; i < m; i++) {scanf("%s", s[i].str);}sort(s, s+m, cmp);for(int i = 0; i < m; i++) {int len = strlen(s[i].str);u = s[i].str[0] - 'a'; v = s[i].str[len-1] - 'a';vis[u] = vis[v] = 1;add(u, v, s[i].str);od[u]++; id[v]++;}bool flag = judge();pi = 0;if(flag) Euler();if(pi != m) puts("***");else {printf("%s", path[pi-1]);for(int i = pi-2; i >= 0; i--) {printf(".%s", path[i]);}puts("");}}return 0;}