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…

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…

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…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1512 题目大意:有n个猴子,一开始每个猴子只认识自己.每个猴子有一个力量值,力量值越大表示这个猴子打架越厉害.如果2个猴子不认识,他们就会找他们认识的猴子中力量最大的出来单挑,单挑不论输赢,单挑的2个猴子力量值减半,这2拨猴子就都认识了,不打不相识嘛.现在给m组询问,如果2只猴子相互认识,输出-1,否则他们各自找自己认识的最牛叉的猴子单挑,求挑完后这拨猴子力量最大值. /* 每次给出要争吵的猴子a和…

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…