并查集+左偏树.....合并的时候用左偏树,合并结束后吧父结点全部定成树的根节点,保证任意两个猴子都可以通过Find找到最厉害的猴子                       Monkey King Time Limit: 10000MS   Memory Limit: 32768KB   64bit IO Format: %lld & %llu [Submit]   [Go Back]   [Status] Description Once in a forest, there lived…

Monkey King Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 4714    Accepted Submission(s): 2032 Problem Description Once in a forest, there lived N aggressive monkeys. At the beginning, they e…

题目地址:P1456 Monkey King 一道挺模板的左偏树题 不会左偏树?看论文打模板,完了之后再回来吧 然后你发现看完论文打完模板之后就可以A掉这道题不用回来了 细节见代码 #include <bits/stdc++.h> using namespace std; const int N = 1e5 + 6; int n, m, f[N], a[N], l[N], r[N], d[N]; //类并查集路径压缩 int get(int x) { if (x == f[x]) return…

As everyone known, The Monkey King is Son Goku. He and his offspring live in Mountain of Flowers and Fruits. One day, his sons get n peaches. And there are m monkeys (including GoKu), they are numbered from 1 to m, GoKu’s number is 1. GoKu wants to d…

我们知道如果要我们给一个序列排序,按照某种大小顺序关系,我们很容易想到优先队列,的确很方便,但是优先队列也有解决不了的问题,当题目要求你把两个优先队列合并的时候,这就实现不了了 优先队列只有插入 删除 取数的操作,但是却没有合并两个优先队列的操作. 这也是它的局限所在. 本次要介绍的左偏树拥有优先队列的所有功能,同时它还可以合并操作.  树的复杂度都比较低,一般log(n)就够了,左偏树也是如此,左偏树如果一个个结点暴力插入复杂度最大为nlog(n) 还有一种仿照二叉树的算法,这里不做介绍. …

Monkey King Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6667    Accepted Submission(s): 2858 Problem Description Once in a forest, there lived N aggressive monkeys. At the beginning, they e…

The Monkey King Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 487    Accepted Submission(s): 166 Problem Description As everyone known, The Monkey King is Son Goku. He and his offspring live i…

Problem Description Once in a forest, there lived N aggressive monkeys. At the beginning, they each does things in its own way and none of them knows each other. But monkeys can't avoid quarrelling, and it only happens between two monkeys who does no…

题目描述 在一个森林里住着N(N<=10000)只猴子.在一开始,他们是互不认识的.但是随着时间的推移,猴子们少不了争斗,但那只会发生在互不认识(认识具有传递性)的两只猴子之间.争斗时,两只猴子都会请出他认识的猴子里最强壮的一只(有可能是他自己)进行争斗.争斗后,这两只猴子就互相认识.   每个猴子有一个强壮值,但是被请出来的那两只猴子进行争斗后,他们的强壮值都会减半(例如10会减为5,5会减为2).  现给出每个猴子的初始强壮值,给出M次争斗,如果争斗的两只猴子不认识,那么输出争斗后两只猴子的…

Once in a forest, there lived N aggressive monkeys. At the beginning, they each does things in its own way and none of them knows each other. But monkeys can't avoid quarrelling, and it only happens between two monkeys who does not know each other. A…

https://www.luogu.org/problemnew/show/1456 #include<cstdio> #include<iostream> #include<algorithm> using namespace std; #define N 100001 struct node { int lc,rc; int key,dis; }e[N]; int fa[N]; void read(int &x) { x=; char c=getchar()…

Description Once in a forest, there lived N aggressive monkeys. At the beginning, they each does things in its own way and none of them knows each other. But monkeys can't avoid quarrelling, and it only happens between two monkeys who does not know e…

  题目描述 Once in a forest, there lived N aggressive monkeys. At the beginning, they each does things in its own way and none of them knows each other. But monkeys can't avoid quarrelling, and it only happens between two monkeys who does not know each o…

[pixiv] https://www.pixiv.net/member_illust.php?mode=medium&illust_id=34693563 向大(hei)佬(e)势力学(di)习(tou) 前段时间学的左偏树,今天复习了一遍,写篇博客加深印象. 左偏树是可合并堆中好写好理解的数据结构.其定义为: 1.满足堆的性质 2.节点的左子节点的距离不小于右子节点的距离 3.每一个子树也满足左偏树的性质 如图 然后这个"距离"懵逼了我好久.标准的定义是: 到最近的右儿子为…

题目链接:HDU - 1512 Once in a forest, there lived N aggressive monkeys. At the beginning, they each does things in its own way and none of them knows each other. But monkeys can't avoid quarrelling, and it only happens between two monkeys who does not kn…

ZOJ2334 用左偏树实现优先队列最大的好处就是两个队列合并可以在Logn时间内完成 用来维护优先队列森林非常好用. 左偏树代码的核心也是两棵树的合并! 代码有些细节需要注意. #include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<cmath> #include<vector> #include<algorithm>…

Problem Description Once in a forest, there lived N aggressive monkeys. At the beginning, they each does things in its own way and none of them knows each other. But monkeys can't avoid quarrelling, and it only happens between two monkeys who does no…

struct LeftTree{ int l,r,val,dis; }t[N]; int fa[N]; inline int Find(int x){ return x == fa[x] ? x : fa[x] = Find(fa[x]); } inline int Merge(int x, int y){ if(!x) return y; if(!y) return x; if(t[x].val < t[y].val || (t[x].val == t[y].val && x &g…

[题目分析] 也是堆+并查集. 比起BZOJ 1455 来说,只是合并的方式麻烦了一点. WA了一天才看到是多组数据. 盲人OI (-￣▽￣)- Best OI. 代码自带大常数,比启发式合并都慢 [代码] #include <cstdio> #include <cstring> #include <cstdlib> #include <cmath> #include <set> #include <map> #include <…

原题链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1389 大致题意:N只相互不认识的猴子(每只猴子有一个战斗力值) 两只不认识的猴子之间发生冲突,两只猴子会分别请出它们认识的最强壮的 猴子进行决斗.决斗之后这,两群猴子都相互认识了. 决斗的那两只猴子战斗力减半...有m组询问 输入a b表示猴子a和b发生了冲突,若a,b属于同一个集合输出-1 否则输出决斗之后这群猴子(已合并)中最强的战斗力值... 具体思路:用并查…

左偏树.我是ziliuziliu,我是最强的 #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> #define maxn 100500 using namespace std; int root[maxn],ls[maxn],rs[maxn],val[maxn],father[maxn],n,m; int x,y,di…

数据结构/可并堆 啊……换换脑子就看了看数据结构……看了一下左偏树和斜堆,鉴于左偏树不像斜堆可能退化就写了个左偏树. 左偏树介绍:http://www.cnblogs.com/crazyac/articles/1970176.html 体会:合并操作是可并堆的核心操作(就像LCT里的access),进堆和弹堆顶都是直接调用合并操作实现的. 而合并的实现是一个递归的过程:将小堆与大堆的右儿子合并(这里的大小指的是堆顶元素的大小),直到某个为0.(是不是有点启发式合并的感觉……) 在合并的过程中要维…

左偏树+并查集.左偏树就是可合并二叉堆. /* 1512 */ #include <iostream> #include <string> #include <map> #include <queue> #include <set> #include <stack> #include <vector> #include <deque> #include <algorithm> #include &l…

[题目链接] http://acm.hdu.edu.cn/showproblem.php?pid=1512 [题目大意] 现在有 一群互不认识的猴子,每个猴子有一个能力值,每次选择两个猴子,挑出他们所归属的部落中能力值最强的猴子打架,然后两个最强的猴子能力值减半,之后两个部落就合为一个部落,问每次合并后部落中最强的猴子能力值是多少 [题解] 要求每次取出一堆数字中最大的数字减半再放回去,显然这是优先队列可以完成的操作,但由于之后要将两堆数字合并,所以采用可并优先队列,考虑使用左偏树.在左偏树的维…

题意:两家原始人(猴)打交道后成为一家猴,打交道时两家分别派出最帅的两位猴子,颜值各自减半,问每次打交道后新家族最帅的猴子的颜值.当然,已经是一家子就没有必要打交道了,因为没有猴希望颜值降低,毕竟还得靠这个吃饭. 裸裸的并查集+大根堆: 由于在下写的queue,需要重量优化,不然超时 #include<cstdio> #include<cstdlib> #include<iostream> #include<queue> #include<algori…

欢迎访问~原文出处——博客园-zhouzhendong 去博客园看该题解 题目传送门 - ZOJ2334 题目传送门 - HDU1512 题意概括 在一个森林里住着N(N<=10000)只猴子.在一开始,他们是互不认识的.但是随着时间的推移,猴子们少不了争斗,但那只会发生在互不认识(认识具有传递性)的两只猴子之间.争斗时,两只猴子都会请出他认识的猴子里最强壮的一只(有可能是他自己)进行争斗.争斗后,这两只猴子就互相认识.每个猴子有一个强壮值,但是被请出来的那两只猴子进行争斗后,他们的强壮值都会减…

http://acm.hdu.edu.cn/showproblem.php?pid=1512 题意: 有n只猴子,每只猴子一开始有个力量值,并且互相不认识,现有每次有两只猴子要决斗,如果认识,就不打了,否则的话,这两只都会从它们所认识的猴子中派出一只力量值最大的猴子出来,并且这只猴子的力量值会减半,在打过之后,这两只猴子所在的集体就都认识了. 思路:这是一道左偏树的模板题. 每个节点有5个值,l为左儿子节点,r为右儿子节点,fa为父亲节点(父亲节点的用法与并查集相似),val为优先级(这道题目就…

提交地址:点击打开链接 题意:  N(N<=10^5)仅仅猴子,初始每仅仅猴子为自己猴群的猴王.每仅仅猴子有一个初始的力量值.这些猴子会有M次会面. 每次两仅仅猴子x,y会面,若x,y属于同一个猴群输出-1,否则将x,y所在猴群的猴王的力量值减半,然后合并这两个猴群. 新猴群中力量值最高的为猴王. 输出新猴王的力量值. 分析:涉及集合的查询,合并,取最值. 利用并查集和左偏树就可以解决. #include <cstdio> #include <cstring> #includ…

题目链接:https://www.luogu.org/problemnew/show/P1456 左偏树并查集不加路径压缩吧... #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; const int maxn = 100000 + 10; struct Left_Tree{ int val, fa, son[2…

用并查集维护猴子们的关系,强壮值用左偏树维护就行了 Code #include <cstdio> #include <algorithm> #include <cstring> #define N 100010 using namespace std; int n,fa[N],m,fr[N]; //fr[x]维护猴子x所在集合在左偏树上的编号 int Find(int x){return x==fa[x]?x:fa[x]=Find(fa[x]);} namespace…