How Many Tables Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 17946    Accepted Submission(s): 8822 Problem Description Today is Ignatius' birthday. He invites a lot of friends. Now it's dinn…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1213 简单的并查集 代码: #include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #define maxn 1100 using namespace std; int parent[maxn]; int m,n; int Find(int x) { int s; ;s=pa…

小希的迷宫 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 41540    Accepted Submission(s): 12811 Problem Description 上次Gardon的迷宫城堡小希玩了很久(见Problem B),现在她也想设计一个迷宫让Gardon来走.但是她设计迷宫的思路不一样,首先她认为所有的通道都应该是…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1232 畅通工程 Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 60381    Accepted Submission(s): 32330 Problem Description 某省调查城镇交通状况,得到现有城镇道路统计表,表中列出了每…

-----------刷点水题练习java------------- 题意:给定N点,M边的无向图,问有多少个连通块. 思路:可以搜索;  可以并查集.这里用并查集练习java的数组使用,ans=N,合并一个连通块ans--: 以及函数的调用: 经验1:C++声明数组是int fa[1024]:而java则是int[] fa=new int[1024]; 经验2:不加static是非静态函数,访问需要new出该类的对象来调用,加上static是静态函数 可直接访问或者通过类名访问. import…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1213 Today is Ignatius' birthday. He invites a lot of friends. Now it's dinner time. Ignatius wants to know how many tables he needs at least. You have to notice that not all the friends know each other,…

Today is Ignatius' birthday. He invites a lot of friends. Now it's dinner time. Ignatius wants to know how many tables he needs at least. You have to notice that not all the friends know each other, and all the friends do not want to stay with strang…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1213 Problem Description Today is Ignatius' birthday. He invites a lot of friends. Now it's dinner time. Ignatius wants to know how many tables he needs at least. You have to notice that not all the frie…

题解:1 2,2 3,4 5,是朋友,所以可以坐一起,求最小的桌子数,那就是2个,因为1 2 3坐一桌,4 5坐一桌.简单的并查集应用,但注意题意是从1到n的,所以要减1. 代码: #include <map> #include <set> #include <list> #include <cmath> #include <queue> #include <stack> #include <vector> #includ…

并查集基本知识看:http://blog.csdn.net/dellaserss/article/details/7724401 题意:假设一张桌子可坐无限多人,小明准备邀请一些朋友来,所有有关系的朋友都可以坐同一张桌,没有关系的则要另开一桌,问需要多少张桌子(小明不坐,不考虑小明与其他人的关系)? 思路:常规的并查集.要求出所有人的老大,有几个老大就要几张桌子.那么有关系的都归为同一个老大.用数组实现,再顺便压缩路径. #include <bits/stdc++.h> #define LL…