Play on Words UVA - 10129

最新推荐文章于 2020-02-17 22:35:15 发布

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#UVA - 10129

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本文介绍了一种独特的考古挑战，通过计算机程序解决字谜，以开启隐藏的门。每扇门上都有磁板，上面写有单词，必须按照特定规则排列，使每个单词的首字母与前一个单词的尾字母相同，从而形成连贯的序列。文章详细阐述了输入输出格式，并提供了一个示例，展示了如何使用编程技巧来解决这一复杂问题。

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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 N that indicates the number of plates (1 ≤ N ≤ 100000). Then exactly N lines 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

acm

ibm

acm

malform

mouse

Sample Output

The door cannot be opened.

Ordering is possible.

The door cannot be opened.

#include <bits/stdc++.h>
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
const int maxn = 26;
const int maxm = 1e3+5;
int fa[maxn],idu[maxn],odu[maxn];
char s[maxm];
int i,j;
int T,n,a,b;

int Find(int x)
{
    if(fa[x] == x) return x;
    return fa[x] = Find(fa[x]);
}
int main()
{
    //有向欧拉通路或者有向欧拉回路（即成环）
    //有向欧拉通路条件：（1）两个点得度为奇数其余点均为偶数且入度=出度
    //                  （2）联通图（即只有一个根节点）
    //有向欧拉通路条件：（1）所有点都是入度=出度；
    //                  （2）联通图
    scanf("%d",&T);
    while(T--)
    {
        for(i = 0; i < maxn; i++)
            fa[i] = i;
        mem(idu,0);
        mem(odu,0);
        scanf("%d",&n);
        for(i = 0; i < n; i++)
        {
            scanf("%s",s);
            int len = strlen(s);
            a = s[0]-'a';
            b = s[len-1]-'a';
            idu[a]++;
            odu[b]++;
            int a1 = Find(a),b1 = Find(b);
            if(a1 != b1) fa[a1] = b1;
        }
        int sum = 0,sum2 = 1,cnt = 0;
        for(i = 0; i < maxn; i++)
        {
            if(abs(odu[i]-idu[i])>=2) sum2 = 0;
            if((odu[i]+idu[i])%2 == 1 )
                sum++;
            if((odu[i]||idu[i])&&fa[i]==i)//找根
                cnt++;
        }
        if(cnt==1&&sum2&&(sum == 0||sum == 2) )
            puts("Ordering is possible.");
        else
            puts("The door cannot be opened.");
    }
    return 0;
}