The Monkey King(hdu5201)】的更多相关文章

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…
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…
题目描述 \(m\)个猴子分\(n\)个桃,要求第一个猴子的桃数严格大于其他猴子,问有多少种分法对\(1e9+7取模(\%1e9+7)\) Input \(1≤T≤25 ,1≤n,m≤100000\) 第一行是\(T\),之后\(T\)行,输入\(n\),\(m\) Output 输出每组数据的分发数 Sample Input 2 2 2 3 5 Sample Output 1 5 先来看一下基础的版本,即\(m\)个猴子分\(n\)个桃子,这我们可以利用隔板法解决, 答案就是\(C(m+n-1…
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…