题目传送门

解题思路:

一道欧拉回路的模板题,详细定理见大佬博客,任意门

AC代码:

 1 #include<cstdio>
 2 #include<iostream>
 3 
 4 using namespace std;
 5 
 6 int ans[100000],du[106],n,k = 0x7f7f7f,b[103][103],cnt = 0x3f,bj,tot;
 7 string l;
 8 
 9 inline void dfs(int x) {
10     for(int i = 0;i < 58; i++)
11         if(b[x][i]) {
12             b[x][i] = b[i][x] = 0;
13             dfs(i);
14         }
15     ans[++tot] = x;
16 }
17 
18 int main()
19 {
20     scanf("%d",&n);
21     for(int i = 1;i <= n; i++) {
22         int x,y;
23         cin >> l;
24         x = l[0] - 'A';
25         y = l[1] - 'A';
26         k = min(k,min(x,y));
27         b[x][y] = b[y][x] = 1;
28         du[x]++;
29         du[y]++;
30     }
31     for(int i = 0;i < 58; i++)
32         if(du[i] % 2 == 1) {
33             bj++;
34             cnt = min(cnt,i);
35         }
36     if(bj == 0) dfs(k);
37     else if(bj == 2) dfs(cnt);
38     else {
39         printf("No Solution");
40         return 0;
41     }
42     for(int i = tot;i >= 1; i--)
43         printf("%c",ans[i] + 'A');
44     return 0;
45 }