UVa10129 Play on Words (欧拉路径)

Posted 2020-08-05

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 }