【HDU1116】Play on Words（欧拉路径+并查集）

题目链接

Play on Words Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 9653    Accepted Submission(s): 3313   Problem 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.

 

解题思路：

将每个单词的首尾字母看做一条有向连线，用并查集判断该图是否连通，若不连通则直接输出“The door cannot be opened.”。

反之，则根据有向图的每个顶点出入度相等 或 有且仅有一个顶点出度=入度+1，另一个入度=出度+1，判断是否能够形成欧拉路径，如果可以则输出“Ordering is possible. ”，反之“The door cannot be opened.”。

 

代码：

#include<bits/stdc++.h>
using namespace std;
int pre[30],in[30],out[30],v[30];
int find(int x)
{
	int r=x;
	while(pre[r]!=r){
		r=pre[r];
	}
	return r;
}
void join(int x,int y)
{
	int i=find(x),j=find(y);
	if(i!=j){
		pre[j]=i;
	}
}
int main()
{
	int repeat;
	scanf("%d",&repeat);
	while(repeat--){
		int n,x,y,root=0,num2=0,num3=0,flag=0;
		char s[1005];
		scanf("%d",&n);
		memset(v,0,sizeof(v));
		memset(pre,0,sizeof(pre));
		memset(in,0,sizeof(in));
		memset(out,0,sizeof(out));
		for(int i=0;i<30;i++)pre[i]=i;
		while(n--){
			scanf("%s",s);
			x=s[0]-'a';
			y=s[strlen(s)-1]-'a';
			out[x]++;
			in[y]++;
			join(x,y);
			v[x]=1;v[y]=1;
		}
		for(int i=0;i<30;i++){
			if(v[i] && pre[i]==i)root++;
		}
		if(root>1)printf("The door cannot be opened.\n");
		else{
			for(int i=0;i<30;i++){
				if(v[i]){
					if(in[i]!=out[i]){
						if(in[i]==out[i]-1)num2++;
						else if(in[i]==out[i]+1)num3++;
						else flag=1;
					}
				}
			}
			if((!flag && num2==0 && num3==0) || (!flag && num2==1 && num3==1))
				printf("Ordering is possible.\n");
			else printf("The door cannot be opened.\n");
		}
	}
	return 0;
}