B - Is It A Tree?
Time Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64u
Submit
Status
Practice
POJ 1308
Description
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.
Output
For each test case display the line "Case k is a tree." or the line "Case k is not a tree.", where k corresponds to the test case number (they are sequentially numbered starting with 1).
题目意思: 判断给定的书不是一棵树:满足:1、联通 2 、根结点只有一个(入度为0的结点只有一个)
代码:
#include<stdio.h>
#include<string.h>
int in[1005];
bool used[1005];
int main()
{
int t=1,i,a,b,x,y;
while(scanf("%d%d",&a,&b),a>=0&&b>=0)
{
if(a==0&&b==0)
{
printf("Case %d is a tree.\n",t++);
continue;
}
memset(used,0,sizeof(used));
memset(in,0,sizeof(in));
in[b]++;
used[a]=true;
used[b]=true;
while(scanf("%d%d",&x,&y),x!=0&&y!=0)
{
used[x]=true;
used[y]=true;
in[y]++;
}
int count=0,flag=0;
for(i=1;i<1005;i++)
{
if(used[i])
{
if(in[i]>1)
{
printf("Case %d is not a tree.\n",t++);
flag=1;
break;
}
else if(in[i]==0)
count++;
}
}
if(!flag)
{
if(count!=1)
printf("Case %d is not a tree.\n",t++);
else
printf("Case %d is a tree.\n",t++);
}
}
return 0;
}