欧拉回路
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 18576    Accepted Submission(s): 7219

Problem Description
欧拉回路是指不令笔离开纸面，可画过图中每条边仅一次，且可以回到起点的一条回路。现给定一个图，问是否存在欧拉回路？
 

Input
测试输入包含若干测试用例。每个测试用例的第1行给出两个正整数，分别是节点数N ( 1 < N < 1000 )和边数M；随后的M行对应M条边，每行给出一对正整数，分别是该条边直接连通的两个节点的编号（节点从1到N编号）。当N为0时输入结
束。
 

Output
每个测试用例的输出占一行，若欧拉回路存在则输出1，否则输出0。
 

Sample Input
3 3
1 2
1 3
2 3
3 2
1 2
2 3
0
 

Sample Output
1
0

 #include <map>

 #include <queue>

 #include <cmath>

 #include <vector>

 #include <string>

 #include <cstdio>

 #include <cstring>

 #include <climits>

 #include <iostream>

 #include <algorithm>

 #define INF 0x3f3f3f3f

 using namespace std;

 const int my_max = ;

 int n, m, my_pre[my_max], a, b, my_node[my_max];

 int my_find(int x)

 {

     int n = x;

     while (n != my_pre[n])

         n = my_pre[n];

     int i = x, j;

     while (n != my_pre[i])

     {

         j = my_pre[i];

         my_pre[i] = n;

         i = j;

     }

     return n;

 }

 bool is_eulerian()

 {

     for (int i = ; i <= n; ++ i)

         if (my_node[i] & ) return false;

     int temp = my_find();

     for (int i = ; i <= n; ++ i)

         if (my_find(i) != temp) return false;

     return true;

 }

 int main()

 {

     while (scanf("%d", &n), n)

     {

         scanf("%d", &m);

         memset(my_node, , sizeof(my_node));

         for (int i = ; i <= n; ++ i)

             my_pre[i] = i;

         while (m --)

         {

             scanf("%d%d", &a, &b);

             my_node[a] ++, my_node[b] ++;

             int n1 = my_find(a), n2 = my_find(b);

             my_pre[n1] = n2;

         }

         if (is_eulerian()) printf("1\n");

         else printf("0\n");

     }

     return ;

 }