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

 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 }