poj1386Play on Words(欧拉通路)

题目： 给你多个单词,问你能不能将所有单词组成这样一个序列:序列的前一个单词的尾字母与后一个单词的头字母相同.

思路：将单词看成有向边，字母看成结点，转化为判断有向图是否有欧拉路。必须满足弱连通，全部顶点度数相同或者有两个点一个入度比出度大一，一个出度比入度大一

#include <cstdio>
#include <queue>
#include <cstring>
#include <iostream>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <map>
#include <string>
#include <set>
#include <ctime>
#include <cmath>
#include <cctype>
using namespace std;
#define maxn 100000+10
#define LL long long
int cas=1,T;
int pre[maxn];
int in[maxn];
int out[maxn];
int Find(int x)
{
return pre[x]==-1?x:pre[x]=Find(pre[x]);
}
int main()
{
//freopen("in","r",stdin);
scanf("%d",&T);
while (T--)
{
int n;
scanf("%d",&n);
memset(pre,-1,sizeof(pre));
memset(in,0,sizeof(in));
memset(out,0,sizeof(out));
for (int i = 0;i<n;i++)
{
char str[1005];
scanf("%s",&str);
int u = str[0]-'a';
int v = str[strlen(str)-1]-'a';
in[u]++;
out[v]++;
u = Find(u);
v = Find(v);
if (u!=v)
pre[u]=v;
}
int cnt = 0;
for (int i = 0;i<26;i++)
if ((in[i] || out[i]) && Find(i)==i)
cnt++;
if (cnt>1)
{
puts("The door cannot be opened.");
continue;
}
int c1=0;
int c2=0;
int c3=0;
for (int i = 0;i<26;i++)
{
if (in[i]==out[i])
continue;
else if (in[i]-out[i]==1)
c2++;
else if (out[i]-in[i]==1)
c3++;
else
c1++;
}
if (((c2==0&&c3==0) || (c2==1&&c3==1)) && c1==0)
puts("Ordering is possible.");
else
puts("The door cannot be opened.");
}
//printf("time=%.3lf",(double)clock()/CLOCKS_PER_SEC);
return 0;
}

Description

Some of the secret doors contain a very interesting word puzzle. The team of archaeologists has to solve it to open that doors. Because there is no other way to open the doors, the puzzle is very important for us.

There is a large number of magnetic plates on every door. Every plate has one word written on it. The plates must be arranged into a sequence in such a way that every word begins with the same letter as the previous word ends. For example, the word ``acm''
can be followed by the word ``motorola''. Your task is to write a computer program that will read the list of words and determine whether it is possible to arrange all of the plates in a sequence (according to the given rule) and consequently to open the door.

Input

The input consists of T test cases. The number of them (T) is given on the first line of the input file. Each test case begins with a line containing a single integer number Nthat indicates the number of plates (1 <= N <= 100000). Then exactly Nlines follow,
each containing a single word. Each word contains at least two and at most 1000 lowercase characters, that means only letters 'a' through 'z' will appear in the word. The same word may appear several times in the list.

Output

Your program has to determine whether it is possible to arrange all the plates in a sequence such that the first letter of each word is equal to the last letter of the previous word. All the plates from the list must be used, each exactly once. The words mentioned
several times must be used that number of times.

If there exists such an ordering of plates, your program should print the sentence "Ordering is possible.". Otherwise, output the sentence "The door cannot be opened.".

Sample Input

3
2
acm
ibm
3
acm
malform
mouse
2
ok
ok

Sample Output

The door cannot be opened.
Ordering is possible.
The door cannot be opened.