Luogu 1341 无序字母对 - 欧拉路径
Solution
找一条字典序最小的欧拉路径。
用 $multiset$ 存储领接表。
Code
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #include<set> 5 using namespace std; 6 7 const int N = 1e3 + 5; 8 9 int n, up = 'z' - 'A' + 1; 10 int ans[N], tot, d[N], num; 11 12 multiset<int> to[N]; 13 14 void dfs(int u) { 15 multiset<int> :: iterator i; 16 for (i = to[u].begin(); i != to[u].end(); i = to[u].begin()) { 17 int nt = *i; 18 to[nt].erase(to[nt].find(u)); 19 to[u].erase(to[u].find(nt)); 20 dfs(nt); 21 } 22 ans[++tot] = u; 23 } 24 25 int main() 26 { 27 scanf("%d", &n); 28 for (int i = 1; i <= n; ++i) { 29 char s[3]; 30 scanf("%s", s); 31 to[s[0] - 'A' + 1].insert(s[1] - 'A' + 1); 32 to[s[1] - 'A' + 1].insert(s[0] - 'A' + 1); 33 d[s[1] - 'A' + 1]++; 34 d[s[0] - 'A' + 1]++; 35 } 36 for (int i = 1; i <= up; ++i) 37 if (d[i] & 1) 38 num++; 39 if (num == 1 || num > 2) 40 return puts("No Solution"), 0; 41 if (num == 0) { 42 for (int i = 1; i <= up; ++i) 43 if (d[i]) {dfs(i); break;} 44 } 45 else 46 for (int i = 1; i <= up; ++i) 47 if (d[i] && (d[i] & 1)) {dfs(i); break;} 48 for (int i = tot; i; i--) 49 putchar(ans[i] + 'A' - 1); 50 puts(""); 51 }