并查集的经典题目: CSUOJ 1601: War Time Limit: 1 Sec  Memory Limit: 128 MBSubmit: 247  Solved: 70[Submit][Status][Web Board] Description AME decided to destroy CH’s country. In CH’ country, There are N villages, which are numbered from 1 to N. We say two vill…

In order to strengthen the defense ability, many stars in galaxy allied together and built many bidirectional tunnels to exchange messages. However, when the Galaxy War began, some tunnels were destroyed by the monsters from another dimension. Then m…

Connections in Galaxy War In order to strengthen the defense ability, many stars in galaxy allied together and built many bidirectional tunnels to exchange messages. However, when the Galaxy War began, some tunnels were destroyed by the monsters from…

Connections in Galaxy War Time Limit: 3 Seconds      Memory Limit: 32768 KB 题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3261 Description: In order to strengthen the defense ability, many stars in galaxy allied together and built…

Description In order to strengthen the defense ability, many stars in galaxy allied together and built many bidirectional tunnels to exchange messages. However, when the Galaxy War began, some tunnels were destroyed by the monsters from another dimen…

题目链接:https://vjudge.net/problem/ZOJ-3261 In order to strengthen the defense ability, many stars in galaxy allied together and built many bidirectional tunnels to exchange messages. However, when the Galaxy War began, some tunnels were destroyed by th…

参考链接: http://www.cppblog.com/yuan1028/archive/2011/02/13/139990.html http://blog.csdn.net/roney_win/article/details/9473225 题意:N个星球,编号从0到N-1.每个星球有一个战斗力power,且这N个星球之间建有一些通道,可以相互联系,在星球大战中,一些星球要向和自己联通的星球中power最强且大于自己的星球求救,且在星球大战中会有一些通道被损坏. 两种操作:破坏a和b之间的…

题意:有N个星球,每个星球有自己的武力值.星球之间有M条无向边,连通的两个点可以相互呼叫支援,前提是对方的武力值要大于自己.当武力值最大的伙伴有多个时,选择编号最小的.有Q次操作,destroy为切断连接两点的边,query为查询某星球能不能向它人呼叫支援. 还是需要离线逆向并查集求解.思路和HDU 4496很相似,但是此处不一定是把所有边都删去,所以需要删边的情况建立出最终的状态.因为N可以到1e4,所以可以用map嵌套map的方式记录过程中被删去的边,最后再根据删除情况建立状态. 在合并时需…

//思路详见课本 P 214 页 思路:直接用并查集,set [ k ]  存 k 的朋友所在集合的代表元素,set [ k + n ] 存 k  的敌人 所在集合的代表元素. #include<iostream> #include<cstring> using namespace std; const int maxn=2 *10000 +100; int set[maxn]; int n; int set_find(int d) { if(set[d]<0) return…

题目链接:https://cn.vjudge.net/problem/ZOJ-3261 题意 有n个星星,之间有m条边 现一边询问与x星连通的最大星的编号,一边拆开一些边 思路 一开始是真不会,甚至想用dfs模拟 最后查了一下,这个题原来是要离线操作,拆边就变为合并 这很为难哈哈,本以为有个什么更好的数据结构(动态树?) 存边我们用一个set<int>来存一个数字即可(bfs这类写多了就很容易考虑到压缩数据) 还有一个重要的点,就是并查集的join可以用来维护一个最大(小)数据作为跟节点的值…