Is It A Tree?
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 17520 Accepted Submission(s): 3939
Problem 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.
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
The input will consist of a sequence of descriptions (test cases) followed by a pair of negative integers. Each test case will consist of a sequence of edge descriptions followed by a pair of zeroes Each edge description will consist of a pair of integers;
the first integer identifies the node from which the edge begins, and the second integer identifies the node to which the edge is directed. Node numbers will always be greater than zero.
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).
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.
Source
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这道题 我已经无力吐槽了、尽管和这道题（hdu1272）非常像。可也仅仅能是非常像、、这道题有方向。就这些。
1.0 0也是一棵树
2.不是-1 -1结束，而是两个数同一时候小于0结束，否则会导致TLE
3.我用并查集推断是否成环，然后推断顶点数是否等于边数+1。但是wa了。
。
4.最后我没实用并查集。不过推断出度和顶点数是否等于边数+1，A了
5.在这里我推断边数用的set容器。
由于在set容器里面假设有反复的数字，会仅仅记一次
我不说了。
看代码把、o(︶︿︶)o 唉
#include<stdio.h>
#include <string.h>
#include <set>
using namespace std;
int fa[1005],ru[1005];
int main()
{
set<int>s;
int a,b,edgs,t=1;
while(scanf("%d %d",&a,&b)!=EOF)
{
if(a<0&&b<0)
break;
if(a==0&&b==0)
{
printf("Case %d is a tree.
",t++);
continue;
}
memset(ru,0,sizeof(ru));
ru[b]++;
s.clear();//一定要记得
edgs=1;
s.insert(a),s.insert(b);
while(1)
{
scanf("%d %d",&a,&b);
if(a==0&&b==0)
break;
edgs++,ru[b]++;
s.insert(a),s.insert(b);
}
int flag=1;
for(int i=0;i<1005;i++)
if(ru[i]>1)
flag=0;
if(s.size()==edgs+1&&flag)
printf("Case %d is a tree.
",t++);
else
printf("Case %d is not a tree.
",t++);
}
return 0;
}