一个长度13的尺子,如果在1位置刻点可以量出1和12,13三种刻度.那么至少刻几个点,可以直接量出1-13所有的长度,分别刻在哪几个位置? 注:必须是直接量.即在尺子上能找出一个1-13任意的整数长度. 写了个没什么技术含量的dfs暴力求解.一个可行解是 1, 2, 6, 10. #include <iostream> #include <vector> #include <unordered_map> using namespace std; class Soluti…

题目链接: option=com_onlinejudge&Itemid=8&page=show_problem&problem=1668">https://icpcarchive.ecs.baylor.edu/index.php? option=com_onlinejudge&Itemid=8&page=show_problem&problem=1668 option=com_onlinejudge&Itemid=8&cate…

 Minimal coverage  The Problem Given several segments of line (int the X axis) with coordinates [Li,Ri]. You are to choose the minimal amount of them, such they would completely cover the segment [0,M]. The Input The first line is the number of test…

Minimal coverage The Problem Given several segments of line (int the X axis) with coordinates [Li,Ri]. You are to choose the minimal amount of them, such they would completely cover the segment [0,M]. The Input The first line is the number of test ca…

Sparse Graph Problem Description   In graph theory, the complement of a graph G is a graph H on the same vertices such that two distinct vertices of H are adjacent if and only if they are notadjacent in G. Now you are given an undirected graph G of N…

http://acm.hdu.edu.cn/showproblem.php?pid=5876 Sparse Graph Problem Description   In graph theory, the complement of a graph G is a graph H on the same vertices such that two distinct vertices of H are adjacent if and only if they are not adjacent in…

可以说是区间覆盖问题的例题... Note: 区间包含+排序扫描: 要求覆盖区间[s, t]; 1.把各区间按照Left从小到大排序,如果区间1的起点大于s,则无解(因为其他区间的左起点更大):否则选择起点在s的最长区间; 2.选择区间[li, ri]后,新的起点应更新为ri,并且忽略所有区间在ri之前的部分:  Minimal coverage  The Problem Given several segments of line (int the X axis) with coordinat…

Abstract 问题: 1)evaluation is often not suffcient 2)biased for certain types of algorthms 3)datasets do not have common ground-truth object positions or extents 4)the initial conditions or parameters of the evaluated tracking algorithms are not the sa…

一.实验目的 1.掌握建立MFC应用程序的方法: 2.掌握映射模式. 二.实验内容 1.在MFC中绘制直尺,直尺需要有刻度,类似于日常学生使用的透明塑料直尺,需要建立四个直尺,分别分布在屏幕客户区的上.下.左.右四个边界.尺子需要有刻度,那客户区上端的尺子距离,应该有厘米.5毫米.1毫米刻度,刻度用竖线显示,长度分别为7毫米.6毫米.5毫米,外观类似于学生直尺,右端留出一公分,防止4个尺子碰在一起. 2.画出一坐标系,给出x坐标变化范围.y坐标变化范围,画出坐标轴,并在坐标轴上标出刻度.原点,要…

题目链接:点击打开链接 题意: 输出n l x y 有一根直尺长度为l 上面有n个刻度. 以下n个数字是距离开头的长度(保证第一个数字是0,最后一个数字是l) 要使得 直尺中存在某2个刻度的距离为x . 某2个刻度的距离为y 要加入最少几个刻度. 问: 最少的刻度个数 输出标记的位置. 思路: 分类讨论一下.. 若本身尺子里就有x.y就输出0 若仅仅有x 或仅仅有y就输出一个刻度. 若2个都没有就: 1.加1个刻度ans.这个ans是距离某个刻度距离为x的,然后看一下是否有距离ans为y的刻度,…