POJ 1308 Is It A Tree? (并查集)
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 24237 | Accepted: 8311 |
Description
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
Output
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
- 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 ...
- 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 ...
- 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 ...
- POJ 1308 Is It A Tree?
Is It A Tree? Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 18778 Accepted: 6395 De ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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
- C:常量、变量 、 表达式 、运算符、 枚举
常量 变量 表达式 运算符 枚举 1.布尔类型只有真和假 2运算符 >,<,<=,>=,==,!=.判断两个数是否相等要使用双等号‘==’.逻辑运算符的表达式结果非真即假,&a ...
- js 原型的内存分析
使用构造器的弊端:http://www.cnblogs.com/a757956132/p/5258897.html 示例 将行为设置为全局的行为,如果将所有的方法都设计为全局函数的时候, 这个函数就可 ...
- 数据库相关文章转载(1) MySQL性能优化之参数配置
1.目的: 通过根据服务器目前状况,修改Mysql的系统参数,达到合理利用服务器现有资源,最大合理的提高MySQL性能. 2.服务器参数: 32G内存.4个CPU,每个CPU 8核. 3.MySQL目 ...
- C++学习笔记之this指针
为了说明这个问题,首先来建立一个简单的类 #include <iostream> #include <string> using namespace std; class Bo ...
- Android开发 Unity3D基础 Android Development
开发环境 Window 7 Unity3D 3.3.0 MB525 defy Android 2.1-update1 本次学习: 1.认识Unity 2.Unity3D环境搭建与Android软件生成 ...
- 大型JavaScript应用程序架构模式
11月中旬在伦敦举行的jQuery Summit顶级大会上有个session讲的是大型JavaScript应用程序架构,看完PPT以后觉得甚是不错,于是整理一下发给大家共勉. PDF版的PPT下载地址 ...
- 高级Swing——列表
1. 列表 1.1 JList构件 JList可以将多个选项放置在单个框中.为了构建列表框,首先需要创建一个字符串数组,然后将这个数组传递给JList构造器. String[] words= { &q ...
- Apache Shiro 使用手册---转载
原文地址:http://www.360doc.com/content/12/0104/13/834950_177177202.shtml (一)Shiro架构介绍 一.什么是Shiro Apache ...
- 21 Free SEO Tools For Bloggers--reference
http://dizyne.net/21-free-seo-tools-for-bloggers/ What do you think is important in a website? Yes, ...
- mysql避免插入重复数据
我们在进行数据库操作的时候,有时候需要插入不重复的数据.所谓不重复的数据,可以是某个字段不重复,也可以是某几个字段重复.当然我们可以在插入之前先将数据库的数据查询出来,然后与将要插入的数据进行对比,如 ...