POJ 2230 Watchcow
题意:反正就是让你从1开始输出欧拉回路的路径
题解:我也不是特别理解的套圈法
1 #include<cstring> 2 #include<iostream> 3 using namespace std; 4 const int N = 20000+5; 5 int head[N]; 6 int cnt = 0, n, m; 7 struct Node 8 { 9 int nx; 10 int to; 11 bool vis; 12 }Edge[N*5]; 13 void add_edge(int u, int v) 14 { 15 Edge[cnt].to = v; 16 Edge[cnt].nx = head[u]; 17 Edge[cnt].vis = 1; 18 head[u] = cnt++; 19 } 20 void dfs(int u) 21 { 22 for(int i = head[u]; ~i; i = Edge[i].nx) 23 { 24 if(Edge[i].vis) 25 { 26 Edge[i].vis = 0; 27 dfs(Edge[i].to); 28 } 29 } 30 cout << u << endl; 31 } 32 int main() 33 { 34 ios::sync_with_stdio(false); 35 cin.tie(0); 36 memset(head, -1, sizeof(head)); 37 cin >> n >> m; 38 int u, v; 39 for(int i = 1; i <= m; i++) 40 { 41 cin >> u >> v; 42 add_edge(u, v); 43 add_edge(v, u); 44 } 45 dfs(1); 46 return 0; 47 }