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.
#include <iostream>
#include <string>
#include<cstring>
#include<cstdio>
#include <vector>
#include <algorithm>
using namespace std;
const int MAXN = ;
int pre[MAXN],tmp[MAXN],cnt=;
int find(int x)
{
if(pre[x]==-)
return x;
else
return pre[x] = find(pre[x]);
};
int main()
{
int x,y;
bool f = false;
memset(pre,-,sizeof(pre));
int cas=;
while(scanf("%d%d",&x,&y))
{
if(x==-&&y==-) break;
if(x==&&y==)
{
int num = ,ft = find(tmp[]);
if(!f)
{
for(int i=;i<cnt;i++)
{
if(find(tmp[i])!=ft)
num++;
}
if(num) f = true;
}
if(!f)
printf("Case %d is a tree.\n",cas);
else
printf("Case %d is not a tree.\n",cas);
memset(pre,-,sizeof(pre));
cas++;
f = false;
cnt = ;
continue;
}
int fx=find(x),fy=find(y);
tmp[cnt++] = x;
tmp[cnt++] = y;
if(fx!=fy)
pre[fy] = fx;
else
f = true;
}
return ;
}

N - Is It A Tree? 并查集的更多相关文章

  1. 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 ...

  2. Is It A Tree?(并查集)

    Is It A Tree? Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 26002   Accepted: 8879 De ...

  3. CF109 C. Lucky Tree 并查集

    Petya loves lucky numbers. We all know that lucky numbers are the positive integers whose decimal re ...

  4. HDU 5606 tree 并查集

    tree 把每条边权是1的边断开,发现每个点离他最近的点个数就是他所在的连通块大小. 开一个并查集,每次读到边权是0的边就合并.最后Ans​i​​=size[findset(i)],size表示每个并 ...

  5. [Swust OJ 856]--Huge Tree(并查集)

    题目链接:http://acm.swust.edu.cn/problem/856/ Time limit(ms): 1000 Memory limit(kb): 10000 Description T ...

  6. Codeforces Round #363 (Div. 2)D. Fix a Tree(并查集)

    D. Fix a Tree time limit per test 2 seconds memory limit per test 256 megabytes input standard input ...

  7. Is It A Tree?(并查集)(dfs也可以解决)

    Is It A Tree? Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%I64d & %I64u Submi ...

  8. tree(并查集)

    tree Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submis ...

  9. 树上统计treecnt(dsu on tree 并查集 正难则反)

    题目链接 dalao们怎么都写的线段树合并啊.. dsu跑的好慢. \(Description\) 给定一棵\(n(n\leq 10^5)\)个点的树. 定义\(Tree[L,R]\)表示为了使得\( ...

  10. 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)Tot ...

随机推荐

  1. [Usaco2011 Jan]道路和航线

    Description Farmer John正在一个新的销售区域对他的牛奶销售方案进行调查.他想把牛奶送到T个城镇 (1 <= T <= 25,000),编号为1T.这些城镇之间通过R条 ...

  2. 407 Trapping Rain Water II 接雨水 II

    给定一个m x n的矩阵,其中的值均为正整数,代表二维高度图每个单元的高度,请计算图中形状最多能接多少体积的雨水.说明:m 和 n 都是小于110的整数.每一个单位的高度都大于0 且小于 20000. ...

  3. Asp.net:MVC认识

    用MVC框架好长一段时间,发现每天都是写业务代码,不想每天只为了工作而写代码,想把工作中认识的MVC框架,遇到的问题,有时候天天在用,但是不知道里面是什么东西,什么原理,为啥這样写等一系列问题.进行梳 ...

  4. angular6 NG-ZORRO 的使用

    1:关于 NG-ZORRO中使用它自己组件改变样式时得使用样式穿透 “class” :: ng-deep "class"

  5. IBatis的分页研究

    IBatis的分页研究 博客分类: Ibatis学习   摘自: http://cpu.iteye.com/blog/311395 yangtingkun   Oracle分页查询语句 ibaits. ...

  6. 通俗理解LDA主题模型(boss)

    0 前言 看完前面几篇简单的文章后,思路还是不清晰了,但是稍微理解了LDA,下面@Hcy开始详细进入boss篇.其中文章可以分为下述5个步骤: 一个函数:gamma函数 四个分布:二项分布.多项分布. ...

  7. JavaScript判断

    if...else: if...else语句是在指定的条件成立时执行的代码,在条件不成立时执行else后的代码. 语法: if(条件) {条件成立时执行的代码 }else{ 条件不成立的时执行的代码} ...

  8. Erwin 带注释(comment )

    1. Database>Pre & Post Script > Model-level %ForEachTable() { alter TABLE %TableName COMME ...

  9. 类似倒圆角方法输入半径选择实体 kword

    ads_name ename; ads_point adspt; acedInitGet(NULL, TEXT("R")); while (1) { int rc = acedEn ...

  10. CAD绘制一个单行文字(com接口VB语言)

    主要用到函数说明: _DMxDrawX::DrawText 绘制一个单行文字.详细说明如下: 参数 说明 DOUBLE dPosX >文字的位置的X坐标 DOUBLE dPosY 文字的位置的Y ...