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首先将单词看成是连接首字母和尾字母的有向边，然后在建立图的过程当中记录每个字母的入度和出度。当无向图建立好了之后，首先利用队列进行DFS遍历，判断图是否联通，如果图不联通则直接否定，如果图是联通的，那么逐个分析每个顶点的入度和出度只差，只有两种情况可以接收：第一种是所有顶点的出度和入度都相等，第二种情况是仅仅存在两个顶点的出度不等于入度，并且其中一个顶点的出度减去入度为1，另外一个顶点的入度减去出度为1，同时要保证出度与入度之间没有其他差值的情况存在即可，具体实现见如下源代码：

#include<iostream>#include<vector>#include<string>#include<set>#include<stack>#include<queue>#include<map>#include<algorithm>#include<cmath>#include<iomanip>#include<cstring>#include<sstream>using namespace std;int T;int main(){cin >> T;while (T--){int in[26], out[26];memset(in,0,sizeof(in));memset(out,0,sizeof(out));int N;cin >> N;int area[26][26];set<char> st;bool visit[26];memset(visit,false,sizeof(visit));memset(area,0,sizeof(area));for (int i = 0; i < N; i++){string s;cin >> s;st.insert(s[0]);st.insert(s[s.size()-1]);int index1 = s[0] - 'a';int index2 = s[s.size() - 1] - 'a';out[index1]++;in[index2]++;area[index1][index2] = 1;area[index2][index1] = 1;}int start = (*st.begin()) - 'a';queue<int> q;q.push(start);visit[start] = true;int amount = 0;while (!q.empty()){int index = q.front();q.pop();amount++;for (int i = 0; i < 26; i++){if (area[index][i] && !visit[i]){q.push(i);visit[i] = true;}}}if (amount != st.size()){cout << "The door cannot be opened." << endl;}else{int first = 0;int second = 0;int other = 0;for (int i = 0; i < 26; i++){if (out[i] - in[i] == 1) first++;else if (in[i] - out[i] == 1) second++;else if (abs(in[i] - out[i]) != 0) other++;}if ((first == 1 && second == 1 && other == 0) || (first == 0 && second == 0 && other == 0))cout << "Ordering is possible." << endl;else cout << "The door cannot be opened." << endl;}}//system("pause");return 0;}