uva 10596 - Morning Walk

Problem H
Morning Walk
Time Limit
3 Seconds
 
Kamal is aMotashota guy. He has got a new job in Chittagong. So, he has moved to Chittagong fromDinajpur. He was getting fatter in Dinajpur as he had no work in his hand there. So, moving to Chittagong has turned to be a blessing for him. Every morning he takes a walk through the hilly roads of charming city Chittagong. He is enjoying this city very much. There are so many roads in Chittagong and every morning he takes different paths for his walking. But while choosing a path he makes sure he does not visit a road twice not even in his way back home. An intersection point of a road is not considered as the part of the road. In a sunny morning, he was thinking about how it would be if he could visit all the roads of the city in a single walk. Your task is to helpKamal in determining whether it is possible for him or not.
 
Input
Input will consist of several test cases. Each test case will start with a line containing two numbers. The first number indicates the number of road intersections and is denoted byN (2 ≤ N ≤ 200). The road intersections are assumed to be numbered from0 to N-1. The second number R denotes the number of roads (0 ≤ R ≤ 10000). Then there will beR lines each containing two numbers c1 andc2 indicating the intersections connecting a road.
 
Output
Print a single line containing the text “Possible” without quotes if it is possible forKamal to visit all the roads exactly once in a single walk otherwise print “Not Possible”.
 
Sample Input
Output for Sample Input
2 2 0 1 1 0 2 1 0 1
Possible Not Possible
 
Problemsetter: Muhammad Abul Hasan
International Islamic UniversityChittagong
회색 달걀이 아픈 제목은 그의 예시를 보면 방향도가 있는 것 같고, 다른 사람의 문제풀이 보고서를 보면 모두 무방향도라고 한다.
무방향도의 오라 회로 문제, 조건은 연통도이고 모든 노드의 도수는 짝수이다

#include <stdio.h>
#define size 210
int num,f,n,visit[size],map[size][size],degree[size];
int dfs(int x)
{int  i;
 for (i=0;i<n;i++)
 if ((visit[i]==1)&&(map[x][i]==1))
 {visit[i]=0; ++num;
  dfs(i);
 }
};
int main()
{int i,j,r,x,y;
 while (scanf("%d%d",&n,&r)!=EOF)
 {
  for (i=0;i<n;i++)
  for (j=0;j<n;j++)
  map[i][j]=0;
  for (i=0;i<n;i++)
  {visit[i]=1; degree[i]=0;}
  for (i=1;i<=r;i++)
  {scanf("%d%d",&x,&y);
   map[x][y]=1;
   ++degree[x];
   ++degree[y];
  }
  if (r>0) f=1; else f=0;
  for (i=0;i<n;i++)
  if (degree[i]%2==1) {f=0;break;}
  if (f) {num=1; visit[0]=0; dfs(0); // visit[n-1]；dfs（n-1）； ， 0 ， 0 ， 。 .....
          if (num!=n) f=0;
         }
  if (f) printf("Possible
");
    else printf("Not Possible
");
 }
 return 0;
}