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A tree is a well-known data structure that is either empty (null, void, nothing) or is a set of one or more nodes connected by directed edges between nodes satisfying the following properties.
There is exactly one node, called the root, to which no directed edges point.
Every node except the root has exactly one edge pointing to it.
There is a unique sequence of directed edges from the root to each node.
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.
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.
题解：该题我是当作水题来做，没啥好说的。注意点数减去边数等于1和入度不能大于一。
AC 代码如下：
#include<stdio.h>
#include<string.h>
int main()
{
int a[1000],b[1000],x,y,d,i=1,c=0,n=0,bbb=1;
memset(a,-1,sizeof(a));
memset(b,0,sizeof(b));
while(1)
{ scanf("%d%d",&x,&y);
if(x<0||y<0)break;
else if(x==0&&y==0)
{
if(c==0)printf("Case %d is a tree.\n",i++);
else if(bbb&&c==n+1)printf("Case %d is a tree.\n",i++);
else printf("Case %d is not a tree.\n",i++);
memset(a,-1,sizeof(a));
memset(b,0,sizeof(b));
c=0;n=0;bbb=1;
}
else
{
if(bbb)
{
if(a[x]==-1){a[x]=1;c++;}
if(a[y]==-1){a[y]=1;c++;}
n++;
b[y]++;
if(b[y]>1)bbb=0;
}
}
}
return 0;
}
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原文地址：http://www.cnblogs.com/441179572qqcom/p/5693175.html
