USACO 3.3 fence 欧拉回路

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

 //主程序部分
 # circuit is a global array
    find_euler_circuit
      circuitpos =
      find_circuit(node )
 ---------------------------------------------
 # nextnode and visited is a local array
 # the path will be found in reverse order
 //递归函数
   find_circuit(node i)
     if node i has no neighbors then
       circuit(circuitpos) = node i
       circuitpos = circuitpos +
     else
       while (node i has neighbors)
           pick a random neighbor node j of node i
           delete_edges (node j, node i)
           find_circuit (node j)
       circuit(circuitpos) = node i
       circuitpos = circuitpos +
 -------------------------------------------
 最终结果：将circuit()数组倒序输出即可

 /*
 PROB:fence
 LANG:C++
 */
 #include <iostream>
 #include <cstring>
 #include <cstdio>
 using namespace std;
 #define INF 999999

 int dx=INF,dy=,n,f,x,y,p;
 int e[][];
 int t[],seq[];

 void dfs(int x)
 {
     for (int i=dx;i<=dy;i++)
         if (e[x][i]>)
         {
             e[x][i]--;
             e[i][x]--;
             dfs(i);
         }
     p++;
     seq[p]=x;
 }

 int start()
 {
     for (int i=dx;i<=dy;i++)
         if (t[i]%!=)
             return i;
     return ;
 }

 int main()
 {
     freopen("fence.in","r",stdin);
     freopen("fence.out","w",stdout);

     memset(e,,sizeof(e));
     memset(t,,sizeof(t));
     cin>>f;
     for (int i=;i<=f;i++)
     {
         cin>>x>>y;
         e[x][y]++;
         e[y][x]++;
         t[x]++;
         t[y]++;
         if (x<dx)   dx=x;
         if (y<dx)   dx=y;
         if (x>dy)   dy=x;
         if (y>dy)   dy=y;

     }

     x=start();
     //cout<<dx<<" "<<dy<<"--"<<x<<endl;
     p=;
     dfs(x);

     //cout<<p<<endl;
     for (int i=p;i>=;i--)
         cout<<seq[i]<<endl;

     return ;
 }