6 8  5 3  5 2  6 4
5 6  0 0
Case 1 is a tree.

8 1  7 3  6 2  8 9  7 5
7 4  7 8  7 6  0 0
Case 2 is a tree.

3 8  6 8  6 4
5 3  5 6  5 2  0 0
Case 3 is not a tree.
0 0
Case 4 is a tree.

1 1 0 0
Case 5 is not a tree.

1 2 1 2 0 0
Case 6 is not a tree.

1 2 2 3 4 5 0 0
Case 7 is not a tree.

1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 1 0 0
Case 8 is not a tree.

1 2 2 1 0 0
Case 9 is not a tree.

1 2 2 3 3 1 5 6 0 0
Case 10 is not a tree.
-1 -1
请按任意键继续. . .


目前已知所有测试数据都过了，竟然还WA！！！这个T会卡RP的么？？？码子搁这，看那位大牛能检查出问题来：
#include<iostream>
#include<cstring>
#include<queue>
#include<set>
using namespace std;
const int nax=1000;
int a[nax],b[nax],c[nax];
bool matrix[nax/10][nax/10];
int sum;//点的个数
void bfs(int a)
{
	queue<int>q;
	int tmp,i;
	q.push(a);
	while(!q.empty())
	{
		tmp=q.front();
		q.pop();
		for(i=0;i<sum;++i)
		{
			if(matrix[tmp][i])
			{
				matrix[tmp][i]=false;
				q.push(i);
			}
		}
	}
}
int main()
{
	int edge;
	set<int>pzj;
	int m,n;
	bool is,ak;
	int root;//记录根节点
	int k=0,i,j;
	while(true)
	{
		sum=0;
		memset(matrix,false,sizeof(matrix));
		ak=true;
		pzj.clear();
		edge=0;
		is=true;
		i=0;
		memset(a,0,sizeof(a));
		memset(b,0,sizeof(b));
		memset(c,0,sizeof(c));
		scanf("%d%d",&m,&n);
		if(m==-1 && n==-1)
			break;
		if((m+n)==0)
		{
			printf("Case %d is a tree.\n",++k);
			continue;
		}
		matrix[m][n]=true;
		edge++;
		pzj.insert(m);
		pzj.insert(n);
		c[m]++;
		b[n]++;
		while(scanf("%d%d",&m,&n) && (m+n))
		{
			matrix[m][n]=true;
			edge++;
			pzj.insert(m);
			pzj.insert(n);
			c[m]++;
			if(c[m]==1)
				a[i++]=m;//保存父节点并防止重复保存
			b[n]++;
			if(b[n]>1)//出现入度大于1
				is=false;
		}
		if(!is)
		{
			printf("Case %d is not a tree.\n",++k);
			continue;
		}//满足没有节点入度大于1
		int cou=0;
		if(pzj.size()-1!=edge)
		{
			printf("Case %d is not a tree.\n",++k);
			continue;
		}
		else//满足点边关系
		{
			for(j=0;j<i;j++)
				if(b[a[j]]==0)
				{
					cou++;
					root=a[j];
				}
		}
		if(cou==1)//满足只有一个根的条件
		{
			sum=pzj.size();//点的个数
			bfs(root);//判断有无环的情况
			for(i=0;i<sum;++i)
				for(j=0;j<sum;++j)
					if(matrix[i][j])
						goto hoho;
				printf("Case %d is a tree.\n",++k);
		}

		else
hoho:			printf("Case %d is not a tree.\n",++k);

	}
	return 0;
}

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