AYIT-2020 609第十六周周赛 F - 并查集 (判断是不是一棵树)

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
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.
给出几组测试数据（以-1 -1结束），每组测试数据有几组数a，b（以0 0结束）表示从a，b相连，问能否构成·一个树，构成树的条件为 1.只有一个节点，称为根节点，没有定向边指向它 2.除了根节点外，每个节点都有一条指向它的边 3.从根到每个节点有一个独特的有向边序列。
解析：
对于第一个条件 意思就是 只能有一个祖先，对应第三个条件，它的意思就是从一个点到另一个点只能有一条唯一路径，在实现的时候对输入的a，b，在a,b祖先不同时进行一个合并，使他们有相同祖先，然后接着输入，在每一次输入之后都要进行一个if (root(a)==root(b))的判断，如果相等说明之前从a到b已经有一条路径了，这个已经是第二条了，所以不能满足条件，在满足第三个条件的情况下最后判断一下有多少个祖先，如果多与一个不满足条件。

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
int pre[100010],vis[100010];
int root(int x)
{
	if(x!=pre[x])
	pre[x]=root(pre[x]);
	return pre[x];
}
int merge(int x,int y)
{
	int tx=root(x);
	int ty=root(y);
	if(tx!=ty)
	{
		pre[tx]=ty;
		return 1;
	}
	return 0;
}
int main()
{
	int k=1,x,y,i,j,t=0,flag=0;
	while(1)
	{
		int l=0;
		memset(vis,0,sizeof(vis));
		for(int i=1;i<=100010;i++)
		pre[i]=i;
		flag=0;
		while(~scanf("%d%d",&x,&y))
		{
			l++;
			if(x==-1&&y==-1) return 0;
			if(x==0&&y==0) break;
			if(merge(x,y)==0)
			flag=1;
			vis[x]=1;
			vis[y]=1;
		}
		t=0;//祖先个数
		for(int i=1;i<=100010;i++)
		{
			if(vis[i]&&root(i)==i)
			{
				t++;
			}
		}
		if(t!=1&&l>1)
		flag=1;
		if(flag)
		printf("Case %d is not a tree.\n",k++);
		else
		printf("Case %d is a tree.\n",k++);
	}
}