（POJ - 1308）Is It A Tree? (tree)

Topic link ：1308 -- Is It A Tree?

The question ： Multi group input , Give you some edges every time ,(0,0) Is the end of each output flag ,(-1,-1) It is the end mark of the title output , Finally, judge whether the given side is a tree .

This problem is easy to make mistakes , An empty node is also a tree , That is, direct input 0,0 It's just a tree , The second is that nodes are not necessarily continuous , That is to say, the node given to you is 10, But not necessarily 9 This node , So we need to record which nodes appear in the process , Nothing else , Just use the union set to judge

Here is the code ：

#include<cstdio>
#include<iostream>
#include<cstring>
#include<vector>
#include<algorithm>
#include<map>
#include<cmath>
#include<queue>
using namespace std;
const int N=10003;
int fu[N];
bool vis[N];
int find(int x)
{
	if(x==fu[x]) return fu[x];
	return fu[x]=find(fu[x]);
}
int main()
{
	int T=0;
	int u,v;
	while(scanf("%d%d",&u,&v)&&(u!=-1||v!=-1))
	{
		if(u==0&&v==0)
		{
			printf("Case %d is a tree.\n",++T);
			continue;
		}
		u++;v++;
		for(int i=1;i<=5001;i++)
			fu[i]=i,vis[i]=false;
		int mx=0,cnt=1;//cnt Record edges ,mx Record the number of different nodes in the tree  
		if(!vis[u]) vis[u]=true,mx++;
		if(!vis[v]) vis[v]=true,mx++;
		bool flag=true;
		int f1=find(u),f2=find(v);
		if(f1==f2)	flag=false;
		else fu[f2]=f1;
		while(1)
		{
			scanf("%d%d",&u,&v);
			if(u==0&&v==0) break;
			u++;v++;
			if(!vis[u]) vis[u]=true,mx++;
			if(!vis[v]) vis[v]=true,mx++;
			f1=find(u),f2=find(v);
			if(f1==f2)	flag=false;
			else fu[f2]=f1;
			cnt++;
		}
		if(!flag||cnt<mx-1)
			printf("Case %d is not a tree.\n",++T);
		else
			printf("Case %d is a tree.\n",++T);
	}
	return 0;
}