标签：hdu1325   hdu 1325   二叉树   

Is It A Tree?

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. 

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

Case 1 is a tree.
Case 2 is a tree.
Case 3 is not a tree.

这道题 我已经无力吐槽了、虽然和这道题（hdu1272）很像，可也只能是很像、、这道题有方向。就这些。

1.0 0也是一棵树

2.不是-1 -1结束，而是两个数同时小于0结束，否则会导致TLE

3.我用并查集判断是否成环，然后判断顶点数是否等于边数+1。可是wa了。。

4.最后我没有用并查集，仅仅是判断出度和顶点数是否等于边数+1，A了

5.在这里我判断边数用的set容器。因为在set容器里面如果有重复的数字，会只记一次

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.\n",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.\n",t++);
        else
        printf("Case %d is not a tree.\n",t++);
    }
    return 0;
}

hdu1325 Is It A Tree?(二叉树的判断)

标签：hdu1325   hdu 1325   二叉树   

原文地址：http://blog.csdn.net/su20145104009/article/details/47089945