Error Curves Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 4928    Accepted Submission(s): 1867 Problem Description Josephina is a clever girl and addicted to Machine Learning recently. Shepay…

1548: Design road Submit Page    Summary    Time Limit: 2 Sec     Memory Limit: 256 Mb     Submitted: 450     Solved: 237 Description You need to design road from (0, 0) to (x, y) in plane with the lowest cost. Unfortunately, there are N Rivers betwe…

P3382 [模板]三分法 题目描述 如题,给出一个N次函数,保证在范围[l,r]内存在一点x,使得[l,x]上单调增,[x,r]上单调减.试求出x的值. 输入输出格式 输入格式: 第一行一次包含一个正整数N和两个实数l.r,含义如题目描述所示. 第二行包含N+1个实数,从高到低依次表示该N次函数各项的系数. 输出格式: 输出为一行,包含一个实数,即为x的值.四舍五入保留5位小数. 输入输出样例 输入样例#1: 复制 3 -0.9981 0.5 1 -3 -3 1 输出样例#1: 复制 -0.4…

题意: 给你一幅图,要你找一个hotel能够满足出去回来,而且保证权值最小: 思路: 可以搜环,然后取最小权值环,拿个点: floyd方便,初始话自己到自己就是无穷,然后就枚举一下给出的hotel就好了 #include<bits/stdc++.h> using namespace std; const int N=1e2+10; const int INF=0x3f3f3f3f; int ma[N][N],hotels[N]; int n,m,C; void floyd() { for(in…

Error Curves Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 4590    Accepted Submission(s): 1753 Problem Description Josephina is a clever girl and addicted to Machine Learning recently. Shepay…

find the most comfortable road Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5359    Accepted Submission(s): 2327 Problem Description XX 星有许多城市,城市之间通过一种奇怪的高速公路SARS(Super Air Roam Structure---超…

/** 题目:A Simple Nim 链接:http://acm.hdu.edu.cn/showproblem.php?pid=5795 题意:给定n堆石子,每堆有若干石子,两个人轮流操作,每次操作可以选择任意一堆取走任意个石子(不可以为空) 或者选择一堆,把它分成三堆,每堆不为空.求先手必胜,还是后手必胜. 思路: 组合游戏Nim: 计算出每一堆的sg值,然后取异或.异或和>0那么先手,否则后手. 对于每一堆的sg值求解方法: 设:sg(x)表示x个石子的sg值.sg(x) = mex{sg…

With the 2010 FIFA World Cup running, football fans the world over were becoming increasingly excited as the best players from the best teams doing battles for the World Cup trophy in South Africa. Similarly, football betting fans were putting their…

问题描述 Huffman树在编码中有着广泛的应用.在这里,我们只关心Huffman树的构造过程. 给出一列数{pi}={p0, p1, …, pn-1},用这列数构造Huffman树的过程如下: 1. 找到{pi}中最小的两个数,设为pa和pb,将pa和pb从{pi}中删除掉,然后将它们的和加入到{pi}中.这个过程的费用记为pa + pb. 2. 重复步骤1,直到{pi}中只剩下一个数. 在上面的操作过程中,把所有的费用相加,就得到了构造Huffman树的总费用. 本题任务:对于给定的一个数列…

653. 两数之和 IV - 输入 BST 给定一个二叉搜索树和一个目标结果,如果 BST 中存在两个元素且它们的和等于给定的目标结果,则返回 true. 案例 1: 输入: 5 / \ 3 6 / \ \ 2 4 7 Target = 9 输出: True 案例 2: 输入: 5 / \ 3 6 / \ \ 2 4 7 Target = 28 输出: False /** * Definition for a binary tree node. * public class TreeNode {…