Is It A Tree?(并查集)

最新推荐文章于 2025-08-12 21:03:53 发布

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最新推荐文章于 2025-08-12 21:03:53 发布

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文章标签： java 数据结构与算法

原文链接：http://www.cnblogs.com/handsomecui/p/4686446.html

本文介绍了一个经典的图论问题“IsItATree?”，并提供了一种使用并查集来判断给定集合是否构成树的有效算法。通过具体代码实现，展示了如何利用节点之间的连接关系来验证树的定义条件。

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Is It A Tree?

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 17569 Accepted Submission(s): 3945

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.
      

题解：参考了大神的解法，好多地方写的很好，但是要判断入度，别问我为什么，我也不知道，但大神就是这样写的。。。

代码：

 1 #include<stdio.h>
 2 #include<string.h>
 3 const int MAXN=100010;
 4 int tree[MAXN],in[MAXN];//in 入度 
 5 int find(int x){
 6     int r=x;
 7     while(r!=tree[r])
 8         r=tree[r];
 9 //    int i=x,j;
10 //    while(i!=r) j=tree[i], tree[i]=r,i=j;
11     tree[x] = r;//需要的才 压缩路径，会长讲的好棒； 
12     return r;
13 }
14 int main(){
15     int x,y,f1,f2,flot=0,tot=0,circle;
16     while(1){
17     tot++;
18     memset(tree,0,sizeof(tree));
19     memset(in,0,sizeof(in)); 
20     while(scanf("%d%d",&x,&y),x||y){
21     if(x<0&&y<0)break;
22     if(!tree[x])
23         tree[x]=x;//
24     if(!tree[y])
25         tree[y]=y;//这里很好，需要了再让tree【y】=y; 
26         in[y]++;//大神的入度，不理解；；； 
27         f1=find(x);//**刚开始写错了写成tree了，找了半天错误 
28         f2=find(y);//**
29         if(f1!=f2)
30             tree[f1]=f2;
31         else flot=1;
32     }
33     if(x<0&&y<0)break;
34     circle=0;
35     for(int i=1;i<MAXN;++i){
36         if(tree[i]==i)circle++;
37         if(circle>1){
38             flot=1;break;
39         }
40         if(in[i]>1){
41             flot=1;break;
42         }
43     }//printf("circle=%d\n",circle);
44     if(flot)printf("Case %d is not a tree.\n",tot);
45     else printf("Case %d is a tree.\n",tot);
46     flot=0;    
47     }
48     return 0;
49 }

转载于:https://www.cnblogs.com/handsomecui/p/4686446.html