Play on Words UVA - 10129 (欧拉路径)

绝地灬酷狼 2022-02-26 03:14 118阅读 0赞

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

3 acm

malform

mouse

## Sample output ##

The door cannot be opened.

Ordering is possible.

The door cannot be opened.

解题思路:

把首尾字母看成结点,单词看成边,分别构建有向图和无向图

然后判断是否存在欧拉道路:判断无向图是否存在欧拉回路：(1)图联通(dfs或者并查集判图联通)(2)所有顶点的度数均是偶数

判断有向图是否存在欧拉回路：(1)底图(忽略边方向后的无向图)联通(2)每个顶点的入度等于出度

判断无向图是否存在欧拉路径：(1)图联通(dfs或者并查集判图联通)(2)仅有2个顶点(起点和终点)的度数为奇数，其他顶点的度数均是偶数。

判断有向图是否存在欧拉路径：(1)底图(忽略边方向后的无向图)联通(2)只有两个点的入度不等于出度，并且其中一个点出度恰好比入度大1，另一个恰好入度比出度小1，每个顶点其它每个点的入度等于出度

首先判断连同分量,利用DFS+无向图即可判断

然后判断结点处入度数,符合上述条件就Yes,否则No

#include <iostream>
    #include <algorithm>
    #include <vector>
    #include <queue>
    #include <string.h>
    #include <stdio.h>
    #include <string>
    #include <set>
    using namespace std;
    const int MAXN = 310;
    int N,m;
    string inputs;
    vector<char> nnedges[MAXN];  //无向边
    int indegress[MAXN];  //变的度数
    int outdegress[MAXN];
    bool visited[MAXN];
    void DFS_Traverse(int node){
    	visited[node] = true;
    	for(int i = 0; i < nnedges[node].size();++i){
    		if(!visited[(int)nnedges[node][i]]){
    			DFS_Traverse((int)nnedges[node][i]);  //深度优先
    		}
    	}
    }
    int main(){
        scanf("%d",&N);  //测试数据总和
    	for(int i = 0; i < N;++i){
    		set<char> nodes;  
    		memset(nnedges,0,sizeof(nnedges));    //无向边
    		memset(visited,false,sizeof(visited));  //设置为未被访问
    		memset(indegress,0,sizeof(indegress));  //入度清空
    		memset(outdegress,0,sizeof(outdegress));  //出度清空
    		scanf("%d",&m);
    		for(int j = 0; j < m;++j){
    		    cin >> inputs;
    			outdegress[(int)inputs[0]]++;
    			indegress[(int)inputs[inputs.length()-1]]++;   //顶点的度数各+1
    			nodes.insert(inputs[0]),nodes.insert(inputs[inputs.length()-1]);  
    			//无向边
    			nnedges[(int)inputs[0]].push_back(inputs[inputs.length()-1]);
    			nnedges[(int)inputs[inputs.length()-1]].push_back(inputs[0]);
    		}
    		//判断连通性,如果不行,直接退
    		int conn = 0;
    		for(auto itr = nodes.begin();itr != nodes.end();++itr){
    			if(!visited[(int)*itr]){
    			    DFS_Traverse((int)*itr);
    				conn++;
    			}
    		}
    		if(conn > 1){
    			printf("The door cannot be opened.\n");  //不能开门
    			continue;
    		}
    		//判断度数
    		int noteuler = 0;
    		bool positive = false;
    		bool negative = false;
    		for(auto itr = nodes.begin();itr != nodes.end();++itr){
    			if((indegress[*itr] != outdegress[*itr])){
    				noteuler++;
    				if(indegress[*itr] - outdegress[*itr] == 1){
    					positive = true;
    				}
    				if(indegress[*itr] - outdegress[*itr] == -1){
    					negative = true;
    				}
    			}
    		}
    		if(noteuler == 0 ||(noteuler == 2 && positive == true && negative == true)){
    			printf("Ordering is possible.\n"); //可行的情况
    		}else{
    			printf("The door cannot be opened.\n");
    		}
    	}
    
    	system("PAUSE");
    	return 0;
    }