Is It A Tree?

本文深入探讨了AI音视频处理与AR特效技术领域的最新进展,包括视频分割、语义识别、自动驾驶、AR增强现实、SLAM、物体检测识别、语音识别变声等关键技术,并详细介绍了其在AR美颜直播特效、视频剪辑、3D空间视频等场景的应用。同时,文章还涉及了图像处理、音视频编解码、硬件加速、OpenCV等基础技术,为开发者提供了一站式的解决方案和技术指导。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

Is It A Tree?

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 596 Accepted Submission(s): 205
 
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.
 
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
North Central North America 1997
 
Recommend
Ignatius.L

poj上能够通过的代码正确的:

#include<iostream>
using namespace std;
bool istree=true;
int parent[4001];
int node[4001];
void init()
{
   for(int i=1;i<=4000;i++)
   {
     parent[i]=-1;
node[i]=-1;
   }
   istree=true;


}
int find(int a)
{
    while(parent[a]>0)
{
  a=parent[a];
}
return a;


}
void combine(int a,int b,int &head)
{
   int p1=find(a);
   int p2=find(b);
   if(p1==p2)
   {
      istree=false;
 return;
   }
   else{
  if(parent[p1]>=parent[p2])
  {
  head=p2;
  parent[p2]+=parent[p1];
      parent[p1]=p2;
  
         
  }
  else{
  head=p1;
  parent[p1]+=parent[p2];
      parent[p2]=p1;
  
  
  }
 
   
   
   }




}
int main()
{
//freopen("in.txt","r",stdin);


  int a,b;
  int k=0;
  while(scanf("%d%d",&a,&b)&&(a>=00&&b>=0))
  {
 int count=0;
 int head=0;
 k++;
      init();
 int temp=a;
 if(a==0&&b==0)
 {
          printf("Case %d is a tree.\n",k);
 continue;
 }
 combine(a,b,head);
 if(node[a]==-1)
 {
    count++;
node[a]=0;
 }
 if(node[b]==-1)
 {
   count++;
node[b]=0;
 
 }
 cin>>a>>b;
 while(a!=0&&b!=0)
 {
    
   if(istree)
{
  
    if(node[a]==-1)
{
        count++;
    node[a]=0;
}
       if(node[b]==-1)
{
        count++;
    node[b]=0;
 
}
   combine(a,b,head);


}
cin>>a>>b;

 
 }
 int len=parent[head]*(-1);


 if(istree&&len==count)
 {
    printf("Case %d is a tree.\n",k);
 
 
 }
 else{
   printf("Case %d is not a tree.\n",k);
 }
 
  
  
  }
  return 0;


}

hoj这题有错误




评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值