UVA - 10129 Play on Words

Play on Words

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 Specification

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 Specification

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

32acmibm3acmmalformmouse2okok

Output for the Sample Input

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



题意：
我们将每一个字母转化成欧拉回路，每一个字母的头和尾，相互连接，可以任意组合，判断最后能不能链接起来；
判断欧拉回路的条件：

1. 最多只能有两个点的入度不等于出度，
2. 而且最终必须是其中一个点的出度恰好比入度大1（把它作为起点），另一个入度比出度大1（把它作为终点）。
3. 当然前提条件是：在忽略边的方向后，图必须是连通的

 

#include 
#include 
#include 
#include 
#define N 26
#define M 10000
using namespace std;

int vis[N];
int in[N], out[N];
int edge[N][N];
int n, m;

void dfs(int u) {
	vis[u] = 1;
	for (int i = 0; i < N; i++) {	
		if (vis[i] == 0 && edge[u][i])
			dfs(i);
	}

}

int judge() {
	int num = 0;
	for (int i = 0; i < N; i++) {
		if (in[i] != out[i]) {
			if (fabs(in[i] - out[i])== 1)
				num++;
			else
				return 0;

		}
	}
	if (num > 2) 
		return 0;
	else {
		for (int i = 0; i < N; i++) {
			if (out[i]) {
				dfs(i);	
			}
		}
		for (int i = 0; i < N; i++) {
			if (out[i] && vis[i] == 0)
				return 0;
			if (in[i] && vis[i] == 0)
				return 0;
		}
		return 1;
	}
}


int main() {
	char a[M];
	while (scanf("%d",&n) != EOF) {
		while (n--) {
			scanf("%d",&m);
			//getchar();
			memset(in,0,sizeof(in));
			memset(out,0,sizeof(out));
			memset(vis,0,sizeof(vis));
			memset(edge,0,sizeof(edge));
			for (int i = 0; i < m; i++) {
				scanf("%s",a);
				int len = strlen(a);
				int u,v;
				u = a[0] - 'a';
				v = a[len - 1] - 'a';
				edge[u][v]++;
				out[u]++;
				in[v]++;
			}

			int k = judge();
			if (k) 
				printf("Ordering is possible.\n");
			else
				printf("The door cannot be opened.\n");
		}

	}

	return 0;
}