poj 2337 Catenyms（并查集+dfs+欧拉回路）
Catenyms
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 131072/65536K (Java/Other)
Total Submission(s) : 6 Accepted Submission(s) : 5
Problem 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:
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.
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 ***
题目大意：输入n个单词，每个单词都形成一条从该单词首字母到尾字母的边，单词尾字母要与下一个单词首字母相同，若可以组成这样的路，即可以组成这样一条连着的单词串，输出路径（单词串），若有多条，则要按字典顺序输出，找不到路则输出***。
此题搞了很长的时间啊，主要是建边那一部分，看了很久，不能用邻接矩阵，后来就用了链表，还有输出路径这个难搞啊，后来参考了一下别人的代码，发现那个dfs路径竟然如此巧妙！实在很巧妙！
代码：
#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <cstring>
#include <string>
#include <algorithm>
#include <math.h>
using namespace std;
bool app[30];
char res[1005][26];
int in[30], out[30], deg[30], father[30], adj[30];
int n, ant, odd, begin, end;
struct edge
{
bool vis;
char sh[25];
int y, next;
}a[1005];
bool cmp(edge a, edge b)
{
return strcmp(a.sh, b.sh) > 0;
}
void init()
{
int i;
ant = odd = 0;
memset(app, false, sizeof(app));
memset(in, 0, sizeof(in));
memset(out, 0, sizeof(out));
memset(deg, 0, sizeof(deg));
memset(adj, -1, sizeof(adj));
for(i = 0; i < 1005; i++) a[i].vis = false;
for(i = 0; i < 26; i++) father[i] = i;
}
int find(int x)
{
if(x != father[x])
{
father[x] = find(father[x]);
}
return father[x];
}
void Union(int x, int y)
{
father[x] = y;
}
int judge() //判断是否满足欧拉。0不满足，1欧拉回路，2欧拉路
{
int i, k;
for(i = 0; i < 26; i++) //判断有向图欧拉
{
if(!app[i]) continue;
deg[i] = in[i] - out[i];
if(abs(deg[i]) > 1) return 0;
if(deg[i] > 0) begin = i; //起点
if(deg[i] < 0) end = i; //终点
if(deg[i]%2) odd++;
if(odd > 2) return 0;
}
for(i = 0; i < 26; i++) if(app[i]) break;
k = find(i);
for(i = k+1; i < 26; i++) //判断连通性
{
if(!app[i]) continue;
if(k != find(i)) return 0;
}
if(odd == 0) //有欧拉回路
{
for(i = 0; i < 26; i++)
if(app[i]) break;
begin = i;
return 1;
}
return 2;
}
void dfs(int x, int id) //深搜寻找欧拉路径，很巧妙！！！
{
int i;
for(i = adj[x]; i != -1; i = a[i].next)
{
if(!a[i].vis)
{
a[i].vis = true;
dfs(a[i].y, i);
}
}
if(id != -1) strcpy(res[ant++], a[id].sh); //最先进去的肯定是终点
}
int main()
{
int i, x, y, fx, fy, t, len, tar;
scanf("%d", &t);
while(t--)
{
init();
scanf("%d", &n);
for(i = 0; i < n; i++)
{
scanf("%s", a[i].sh);
}
//题目要求是字典顺序，但是我是用前插链表，这时的顺序恰好会相反
sort(a, a+n, cmp); //所以排序时从大到小，这样刚刚会是字典顺序
for(i = 0; i < n; i++)
{
len = strlen(a[i].sh);
x = a[i].sh[0] - 'a';
y = a[i].sh[len-1] - 'a';
in[x]++;
out[y]++;
app[x] = true;
app[y] = true;
/// *****建边*****
a[i].y = y;
a[i].next = adj[x];
adj[x] = i;
/// ***************
fx = find(x);
fy = find(y);
if(fx != fy) Union(fx, fy);
}
tar = judge();
if(tar == 0)
{
printf("***\n");
continue;
}
dfs(begin, -1);
printf("%s", res[ant-1]);
for(i = ant-2; i >= 0; i--)
{
printf(".%s", res[i]);
}
printf("\n");
}
return 0;
}