题意：求给定图的欧拉回路（每条边只走一次）
若欧拉回路存在，图中只可能有0个or2个奇数度的点。
求解时，若有奇数度的点，则必须从该点开始。否则可以从任一点开始
求解过程：dfs
1 //主程序部分 2 # circuit is a global array 3 find_euler_circuit 4 circuitpos = 0 5 find_circuit(node 1) 6 --------------------------------------------- 7 # nextnode and visited is a local array 8 # the path will be found in reverse order 9 //递归函数 10 find_circuit(node i) 11 if node i has no neighbors then 12 circuit(circuitpos) = node i 13 circuitpos = circuitpos + 1 14 else 15 while (node i has neighbors) 16 pick a random neighbor node j of node i 17 delete_edges (node j, node i) 18 find_circuit (node j) 19 circuit(circuitpos) = node i 20 circuitpos = circuitpos + 1 21 ------------------------------------------- 22 最终结果：将circuit()数组倒序输出即可
1 /* 2 PROB:fence 3 LANG:C++ 4 */ 5 #include <iostream> 6 #include <cstring> 7 #include <cstdio> 8 using namespace std; 9 #define INF 999999 10 11 int dx=INF,dy=0,n,f,x,y,p; 12 int e[1100][1100]; 13 int t[1100],seq[1100]; 14 15 void dfs(int x) 16 { 17 for (int i=dx;i<=dy;i++) 18 if (e[x][i]>0) 19 { 20 e[x][i]--; 21 e[i][x]--; 22 dfs(i); 23 } 24 p++; 25 seq[p]=x; 26 } 27 28 int start() 29 { 30 for (int i=dx;i<=dy;i++) 31 if (t[i]%2!=0) 32 return i; 33 return 1; 34 } 35 36 int main() 37 { 38 freopen("fence.in","r",stdin); 39 freopen("fence.out","w",stdout); 40 41 memset(e,0,sizeof(e)); 42 memset(t,0,sizeof(t)); 43 cin>>f; 44 for (int i=1;i<=f;i++) 45 { 46 cin>>x>>y; 47 e[x][y]++; 48 e[y][x]++; 49 t[x]++; 50 t[y]++; 51 if (x<dx) dx=x; 52 if (y<dx) dx=y; 53 if (x>dy) dy=x; 54 if (y>dy) dy=y; 55 56 } 57 58 x=start(); 59 //cout<<dx<<" "<<dy<<"--"<<x<<endl; 60 p=0; 61 dfs(x); 62 63 //cout<<p<<endl; 64 for (int i=p;i>=1;i--) 65 cout<<seq[i]<<endl; 66 67 return 0; 68 }