Is It A Tree?
Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 24237   Accepted: 8311

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. 和HDU 1272一模一样,除了改下输出其他什么都不用改。
 #include <iostream>
#include <cstdio>
#include <string>
#include <queue>
#include <vector>
#include <map>
#include <algorithm>
#include <cstring>
#include <cctype>
#include <cstdlib>
#include <cmath>
#include <ctime>
using namespace std; const int SIZE = ;
int FATHER[SIZE],RANK[SIZE];
vector<int> S; void ini(int);
int find_father(int);
bool unite(int,int);
bool same(int,int);
int main(void)
{
int a,b,count = ;
bool flag; while(scanf("%d%d",&a,&b) && (a != - && b != -))
{
flag = true;
if(a == && b == )
{
printf("Case %d %s\n",++ count,flag ? "is a tree." : "is not a tree.");
continue;
}
S.clear();
S.push_back(a);
S.push_back(b);
ini(SIZE - );
flag = unite(a,b);
while(scanf("%d%d",&a,&b) && (a || b))
{
flag &= unite(a,b);
S.push_back(a);
S.push_back(b);
}
for(int i = ;i < S.size();i ++)
flag &= same(S[],S[i]);
printf("Case %d %s\n",++ count,flag ? "is a tree." : "is not a tree.");
} return ;
} void ini(int n)
{
for(int i = ;i <= n;i ++)
{
FATHER[i] = i;
RANK[i] = ;
}
} int find_father(int n)
{
if(n == FATHER[n])
return n;
return FATHER[n] = find_father(FATHER[n]);
} bool unite(int x,int y)
{
x = find_father(x);
y = find_father(y); if(x == y)
return false;
if(RANK[x] < RANK[y])
FATHER[x] = y;
else
{
FATHER[y] = x;
if(RANK[x] == RANK[y])
RANK[x] ++;
}
return true;
} bool same(int x,int y)
{
return find_father(x) == find_father(y);
}

POJ 1308 Is It A Tree? (并查集)的更多相关文章

  1. hdu 1325 && poj 1308 Is It A Tree?(并查集)

    Description A tree is a well-known data structure that is either empty (null, void, nothing) or is a ...

  2. POJ 1308 Is It A Tree?和HDU 1272 小希的迷宫

    POJ题目网址:http://poj.org/problem?id=1308 HDU题目网址:http://acm.hdu.edu.cn/showproblem.php?pid=1272 并查集的运用 ...

  3. HDU 1325,POJ 1308 Is It A Tree

    HDU认为1>2,3>2不是树,POJ认为是,而Virtual Judge上引用的是POJ数据这就是唯一的区别....(因为这个瞎折腾了半天) 此题因为是为了熟悉并查集而刷,其实想了下其实 ...

  4. POJ 1308 Is It A Tree?

    Is It A Tree? Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 18778   Accepted: 6395 De ...

  5. HDU ACM 1325 / POJ 1308 Is It A Tree?

    Is It A Tree? Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Tot ...

  6. POJ 1308 Is It A Tree?--题解报告

    Is It A Tree? Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 31092   Accepted: 10549 D ...

  7. POJ 1308 Is It A Tree? 解题报告

    Is It A Tree? Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 32052   Accepted: 10876 D ...

  8. POJ 1417 - True Liars - [带权并查集+DP]

    题目链接:http://poj.org/problem?id=1417 Time Limit: 1000MS Memory Limit: 10000K Description After having ...

  9. Hdu.1325.Is It A Tree?(并查集)

    Is It A Tree? Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) To ...

随机推荐

  1. erlang: Programming Rules and Conventions。

    http://www.erlang.se/doc/programming_rules.shtml#HDR33 http://www.erlang.org/eeps/eep-0008.html

  2. 26.怎样在Swift中定义宏?

    Swift 中没有宏定义,苹果建议使用let 或者 get 属性来替代宏定义值.虽然没有#define,但我们仍然可以使用 #if 并配合编译的配置来完成条件编译.下面会列出Swift项目开发中的一些 ...

  3. jquery提示信息 tips

    <%@ page language="java" contentType="text/html; charset=UTF-8" pageEncoding= ...

  4. baseDao 使用spring3+hibernate4方式

    启动异常: java.lang.ClassCastException: org.springframework.orm.hibernate4.SessionHolder cannot be cast  ...

  5. 常见的Unix指令

    ls -1 列出当前目录下的所有内容(文件/文件夹) pwd 显示当前操作的目录 cd   改变当前操作的目录 who 显示当前用户 clear 清屏 mkdir 新建一个目录 touch 新建一个文 ...

  6. angularjs directive学习心得

    一些常见的错误 在angularjs里,创建directive时,directive的名称应该要使用驼峰式,例如myDirective,而在html里要调用它的时候,就不能用驼峰式了,可以用my-di ...

  7. 分布式文件系统HDFS体系

    系列文件列表: http://os.51cto.com/art/201306/399379.htm 1.介绍 hadoop文件系统(HDFS)是一个运行在普通的硬件之上的分布式文件系统,它和现有的分布 ...

  8. C# DataGridView中合并单元格

    /// 合并GridView列中相同的行 /// /// GridView对象 /// 需要合并的列 public static void GroupRows(GridView GridView1, ...

  9. C# RSA和Java RSA互通

    今天调查了C# RSA和Java RSA,网上很多人说,C#加密或者java加密 ,Java不能解密或者C#不能解密 但是我尝试了一下,发现是可以的,下面就是我尝试的代码,如果您有什么问题,我想看看, ...

  10. android仿win8 metro磁贴布局

    代码下载     //更新代码,   这里是更新后的代码 //////////////////////// 1,含一个图片无限滚动的控件,自己实现的 2.可新增删除每个磁贴 3.来个图片吧 ////* ...