Play on Words

题目：

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.

 

题意：

给出一些单词，问是否能组成一个序列，让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：

这道题其实就是问是否存在欧拉通路；

无向图G存在欧拉通路的条件是：G为连通图，并且G只有两个奇度节点或者无奇度节点。

只需判断图是否连通，和奇度节点的数量。

 

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn = 100;

int T,n,G[maxn][maxn],c[maxn],in[maxn],out[maxn];
int vis[maxn];

void dfs(int u)
{
    vis[u]=1;
    for (int v=0; v<26; v++)
        if (!vis[v]&&G[u][v])dfs(v);
}

int main()
{
    char s[2000];
    scanf("%d", &T);
    while (T--)
    {
        memset(vis,1,sizeof(vis));
        memset(G,0,sizeof(G));
        memset(in,0,sizeof(in));
        memset(out,0,sizeof(out));
        scanf("%d",&n);
        for (int i=0; i<n; i++)
        {
            scanf("%s",s);
            int len=strlen(s);
            int r=s[0]-'a';
            int t=s[len-1] - 'a';
            G[r][t] = 1;
            in[t]++;
            out[r]++;
            vis[r] = vis[t] = 0;
        }
        int flag=0;
        int p=0,cnt1=0,cnt2=0,cnt3 = 0;
        for (int i=0; i<26; i++)
        {
            if (out[i]==in[i])
                continue;
            if (out[i]==in[i]+1)
            {
                cnt1++;
                p=i;
            }
            else if(in[i]==out[i]+1)
                cnt2++;
            else
                cnt3++;
        }
        if (cnt3>0)
        {
            printf("The door cannot be opened.\n");
            continue;
        }
        if (cnt1==1&&cnt2==1||cnt1==0&&cnt2==0)
            flag=1;
        else
            flag=0;
        dfs(p);
        for (int i=0; i<26; i++)
            if(!vis[i]) flag = 0;
        if (flag) printf("Ordering is possible.\n");
        else
            printf("The door cannot be opened.\n");
    }
    return 0;
}