标签：c++   uva   

                               uva 10129 Play on Words

Description

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

3
2
acm
ibm
3
acm
malform
mouse
2
ok
ok

Output for the Sample Input

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

解题思路：欧拉路的问题，满足 1、所有点的出度与入度的绝对值和不能大于2；

                                                       2、所有点是联通的、

#include<stdio.h>
#include<math.h>
#include<string.h>
using namespace std;
char ch[1005];
int num[30];
int get(int x) {
	return num[x] != x ? get(num[x]) : x;
}
int main () {
	int T, B[30], F[30];
	scanf("%d", &T);
	while (T--) {
		for (int i = 0; i < 26; i++) {
			num[i] = i;
		}
		memset(B, 0, sizeof(B));
		memset(F, 0, sizeof(F));
		int n;
		scanf("%d", &n);
		for (int i = 0; i < n; i++) {
			scanf("%s", ch);
			int x = ch[0] - 'a';
			int y = ch[strlen(ch) - 1] - 'a';
			B[x]++; //统计各字符串的开头结尾字母出现次数
			F[y]++;
			num[get(x)] = get(y); //使用get函数使点联通
		}
		int temp, flag = 0;
		for (int i = 0; i < 26; i++) {
			if ((B[i] || F[i]) && i == get(i)) { //找到起始点
				temp = i;
				break;
			}
		}
		for (int i = 0; i < 26; i++) {
			flag += fabs(B[i] - F[i]); 
			if ((B[i] || F[i]) && temp != get(i)) { //若不联通则不成立
				flag += 10;
			}
		}
		if (flag <= 2) printf("Ordering is possible.\n"); //各点的入度出度之和不能大于2（头尾节点），否则不成立
		else printf("The door cannot be opened.\n");
	}
	return 0;
}

uva 10129 Play on Words（欧拉路）

标签：c++   uva   

原文地址：http://blog.csdn.net/llx523113241/article/details/43153337