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链接：http://acm.hust.edu.cn/vjudge/problem/19492
分析：
欧拉路径：https://zh.wikipedia.org/wiki/%E4%B8%80%E7%AC%94%E7%94%BB%E9%97%AE%E9%A2%98
只用关注每个单词的首尾字母就够了其它的没有用，从首字母向尾字母连边，看是否能形成一个欧拉路径，用DFS找连通块的方式判断是否是连通图，然后如果每个点的出入度数相等则存在欧拉回路当然就有欧拉路径，否则查看是否存在两个点出入度数不相等，且其中一个点的出入度数相差1，那么另一个点的出入度数差肯定与之互补。
1 #include <cstdio> 2 #include <cstring> 3 #include <vector> 4 using namespace std; 5 6 const int maxn = 1000 + 5; 7 8 int deg[256], vis[256]; 9 vector<int> G[256]; 10 11 void dfs(int u) { 12 vis[u] = 1; 13 for (int i = 0; i < G[u].size(); i++) 14 if (vis[G[u][i]]) continue; 15 else dfs(G[u][i]); 16 } 17 18 int main() { 19 int T; 20 scanf("%d", &T); 21 while (T--) { 22 for (int i = ‘a‘; i <= ‘z‘; i++) deg[i] = 0; 23 for (int i = ‘a‘; i <= ‘z‘; i++) G[i].clear(); 24 int n; 25 scanf("%d", &n); 26 char word[maxn]; 27 for (int i = 0; i < n; i++) { 28 scanf("%s", word); 29 char c1 = word[0], c2 = word[strlen(word) - 1]; 30 deg[c1]++, deg[c2]--; 31 G[c1].push_back(c2), G[c2].push_back(c1); 32 } 33 int cnt = 0; 34 for (int ch = ‘a‘; ch <= ‘z‘; ch++) vis[ch] = 0; 35 for (int ch = ‘a‘; ch <= ‘z‘; ch++) 36 if (vis[ch] == 0 && G[ch].size() > 0) { 37 cnt++; if (cnt > 1) break; 38 dfs(ch); 39 } 40 vector<int> d; 41 for (int ch = ‘a‘; ch <= ‘z‘; ch++) 42 if (deg[ch] != 0) d.push_back(deg[ch]); 43 bool ok = false; 44 if (cnt == 1 && (d.empty() || (d.size() == 2 && (d[0] == 1 || d[0] == -1)))) ok = true; 45 if (ok) printf("Ordering is possible.\n"); 46 else printf("The door cannot be opened.\n"); 47 } 48 return 0; 49 }
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原文地址：http://www.cnblogs.com/XieWeida/p/5831650.html
