HDU——1325 Is there a tree

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 19 Accepted Submission(s) : 3

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Problem Description

For example, consider the illustrations below, in which nodes are represented by circles and edges are represented by lines with arrowheads. The first two of these are trees, but the last is not.

HDU——1325 Is there a tree

In this problem you will be given several descriptions of collections of nodes connected by directed edges. For each of these you are to determine if the collection satisfies the definition of a tree or not.

Input

Output

Sample Input

6 8 5 3 5 2 6 4
5 6 0 0
8 1 7 3 6 2 8 9 7 5
7 4 7 8 7 6 0 0
3 8 6 8 6 4
5 3 5 6 5 2 0 0
-1 -1

Sample Output

Case 1 is a tree.
Case 2 is a tree.
Case 3 is not a tree.

有向树，恶心！！！树的节点数等于边数加 1。。。。

View Code

#include <stdio.h>
#define M 100000+1
int parent[M],flag[M];
int main()
{
int a,b,f,i,n,e,count;
    count=1;
while(1)
    {
         scanf("%d %d",&a,&b);
if(a<0 && b<0) break;
         f=1;
for(i=1;i<M;i++) {parent[i]=-1;flag[i]=0;}
         e = n = 0;
while(a && b)
         {
if(!flag[a])
            {
                 flag[a]=1; n++;
            }
if(!flag[b])
            {
                 flag[b]=1; n++;
            }
            e++;
if(parent[b]==-1)
                parent[b]=a;
else
            {
                f=0;break;
            }
            scanf("%d %d",&a,&b);
         }
while(a && b) { scanf("%d %d",&a,&b);}
if(f && (n==0 || n==e+1) )
        {
               printf("Case %d is a tree.\n",count);
        }
else
        {
            printf("Case %d is not a tree.\n",count);
        }
        count++;
    }
return 0;
}