Counting Intersections】的更多相关文章

传送门:hdu 5862 Counting Intersections 题意:对于平行于坐标轴的n条线段,求两两相交的线段对有多少个,包括十,T型 官方题解:由于数据限制,只有竖向与横向的线段才会产生交点,所以先对横向线段按x端点排序,每次加入一个线段,将其对应的y坐标位置+1,当出现一个竖向线段时,查询它的两个y端点之间的和即为交点个数. 注意点:对x坐标排序是对所有线段端点排序:因为可能出现 “  1-1  “ 这样的情况,所以对于横着的线段,需要进行首尾x坐标处理:我的方法是对于x坐标,先…
Counting Intersections 题目链接: http://acm.split.hdu.edu.cn/showproblem.php?pid=5862 Description Given some segments which are paralleled to the coordinate axis. You need to count the number of their intersection. The input data guarantee that no two se…
HDU 5862 Counting Intersections(离散化+树状数组) 题目链接http://acm.split.hdu.edu.cn/showproblem.php?pid=5862 Description Given some segments which are paralleled to the coordinate axis. You need to count the number of their intersection. The input data guarant…
Counting Intersections Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Problem Description Given some segments which are paralleled to the coordinate axis. You need to count the number of their intersection. The i…
题目链接: http://acm.split.hdu.edu.cn/showproblem.php?pid=5862 Counting Intersections Time Limit: 12000/6000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) 问题描述 Given some segments which are paralleled to the coordinate axis. You need to coun…
Counting Intersections Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1138    Accepted Submission(s): 347 Problem Description Given some segments which are paralleled to the coordinate axis. Y…
传送门:Hdu 5862 Counting Intersections 题意:有n条线段,每一条线段都是平行于x轴或者y轴,问有多少个交点 分析: 基本的操作流程是:先将所有的线段按照横树坐标x按小的优先排序,注意是所有的线段 :(这里是将线段都去掉只保留两个端点) 然后从左到右的顺序经行扫描,遇到横的线段,如果是左端点对应的 yi 便++ , 若是右端点对应的y1便--:  遇到竖直的线段,便统计区间[y1,y2] 的数 , 看到这了是不是有点东西了呢? 如果我是按照x排序的话,两线段若想相交…
题目链接: Counting Intersections Time Limit: 12000/6000 MS (Java/Others)     Memory Limit: 65536/65536 K (Java/Others) Problem Description Given some segments which are paralleled to the coordinate axis. You need to count the number of their intersection…
原题链接 Problem Description Given some segments which are paralleled to the coordinate axis. You need to count the number of their intersection. The input data guarantee that no two segments share the same endpoint, no covered segments, and no segments…
题意 有n条线段,且都平行于坐标轴.对于每条线段,给出两个端点的坐标.问一共有多少个线段的交点. 分析 最最简单的扫描法了.用线段树或者树状数组都可以. 由题目可知,线段只有两种,要么平行于x轴要么平行于y轴.而交点只能是两个不平行的线段产生的. 所有我们以一条平行于x轴的线为扫描线,从下向上扫.先把横坐标进行离散化,然后把平行于y轴的线段拆成上下两个端点.当扫到下端点的时候就在它横坐标+1,当扫到上端点的时候,就在它横坐标-1.对于每一条平行于x轴的线,则将左右端点内的值相加.就酱~ 这里一个…
题意:给你若干个平行于坐标轴的,长度大于0的线段,且任意两个线段没有公共点,不会重合覆盖.问有多少个交点. 析:题意很明确,可是并不好做,可以先把平行与x轴和y轴的分开,然后把平行y轴的按y坐标从小到大进行排序,然后我们可以枚举每一个平行x轴的线段, 我们可以把平行于x轴的线段当做扫描线,只不过有了一个范围,每次要高效的求出相交的线段数目,可以用一个优先队列来维护平行y轴的线段的上坐标, 如果在该平行于x轴的范围就给相应的横坐标加1,这样就很容易想到是用树状数组来维护,然后每次求出最左边和最右边…
Given some segments which are paralleled to the coordinate axis. You need to count the number of their intersection. The input data guarantee that no two segments share the same endpoint, no covered segments, and no segments with length 0. InputThe f…
Counting Intersections Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 51    Accepted Submission(s): 18 Problem Description Given some segments which are paralleled to the coordinate axis. You…
solved 7/11 2016 Multi-University Training Contest 10 题解链接 分类讨论 1001 Median(BH) 题意: 有长度为n排好序的序列,给两段子序列[l1,r1],[l2,r2]构成新的序列,问中间的数字. 思路: 根据不同情况分类讨论即可.时间复杂度O(1). 代码: #include <bits/stdc++.h> const int N = 1e5 + 5; int a[N]; int n, m; int l1, r1, l2, r…
Problem Counting Intersections 题目大意 给定n条水平或竖直的线段,统计所有线段的交点个数. (n<=100000) 解题分析 首先将线段离散化. 然后将所有线段按照横坐标的顺序,按照先插入再查找再删除的顺序进行操作. 对于横线 如果是左端点,则将其纵坐标加1,右端点则减1,对于竖线直接求和就行了. 参考程序 #include <map> #include <set> #include <stack> #include <que…
在上篇,我了解了基数的基本概念,现在进入Linear Counting算法的学习. 理解颇浅,还请大神指点! http://blog.codinglabs.org/articles/algorithms-for-cardinality-estimation-part-ii.html 它的基本处理方法和上篇中用bitmap统计的方法类似,但是最后要用到一个公式: 说明:m为bitmap总位数,u为0的个数,最后的结果为n的一个估计,且为最大似然估计(MLE). 那么问题来了,最大似然估计是什么东东…
来之不易的2017第一发ac http://poj.org/problem?id=2386 Lake Counting Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 31474   Accepted: 15724 Description Due to recent rains, water has pooled in various places in Farmer John's field, which is repr…
ZOJ3944 People Counting ZOJ3939 The Lucky Week 1.PeopleConting 题意:照片上有很多个人,用矩阵里的字符表示.一个人如下: .O. /|\ (.) 占3*3格子,句号“.”为背景.没有两个人完全重合.有的人被挡住了一部分.问照片上有几个人. 题解: 先弄个常量把3*3人形存起来,然后6个部位依次找,比如现在找头,找到一个头,就把这个人删掉(找这个人的各个部位,如果在该部位位置的不是这个人的身体,就不删),删成句号,疯狂找就行了. 代码:…
#include <stdio.h> #include <malloc.h> #define MAX_STACK 10 ; // define the node of stack typedef struct node { int data; node *next; }*Snode; // define the stack typedef struct Stack{ Snode top; Snode bottom; }*NewStack; void creatStack(NewSt…
1004. Counting Leaves (30)   A family hierarchy is usually presented by a pedigree tree. Your job is to count those family members who have no child. Input Each input file contains one test case. Each case starts with a line containing 0 < N < 100,…
Problem Figure 2. The Hamming distance between these two strings is 7. Mismatched symbols are colored red. Given two strings ss and tt of equal length, the Hamming distance between ss and tt, denoted dH(s,t)dH(s,t), is the number of corresponding sym…
Problem A string is simply an ordered collection of symbols selected from some alphabet and formed into a word; the length of a string is the number of symbols that it contains. An example of a length 21 DNA string (whose alphabet contains the symbol…
// uva 11401 Triangle Counting // // 题目大意: // // 求n范围内,任意选三个不同的数,能组成三角形的个数 // // 解题方法: // // 我们设三角巷的最长的长度是c(x),另外两边为y,z // 则由z + y > x得, x - y < z < x 当y = 1时,无解 // 当y = 2时,一个解,这样到y = x - 1 时 有 x - 2个 // 解,所以一共是0,1,2,3....x - 2,一共(x - 2) * (x - 1…
当我们在使用JSONKit处理数据时,直接将文件拉进项目往往会报这两个错“JSONKit   does not support Objective-C Automatic Reference Counting(ARC)”,“ARC forbids Objective-C objects in struct”,这是由于JSONKit库未更新,不支持ARC机制.我们可以参照如下步骤解决:…
有使用JSonKit的朋友,如果遇到“JSonKit does not support Objective-C Automatic Reference Counting(ARC)”这种情况,可参照如下方法: 点击项目根目录->targets->Build Phases->JSONKit.m->添加“-fno-objc-arc”字段,在运行就OK了.…
Lake Counting Time Limit: 1000MS     Memory Limit: 65536K Total Submissions: 17917     Accepted: 9069 Description Due to recent rains, water has pooled in various places in Farmer John's field, which is represented by a rectangle of N x M (1 <= N <=…
Boring Counting Time Limit: 3000ms   Memory limit: 65536K  有疑问?点这里^_^ 题目描述     In this problem you are given a number sequence P consisting of N integer and Pi is the ith element in the sequence. Now you task is to answer a list of queries, for each…
Boring counting Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 98304/98304 K (Java/Others) Total Submission(s): 2811    Accepted Submission(s): 827 Problem Description In this problem we consider a rooted tree with N vertices. The vertices a…
Counting Squares Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 1885    Accepted Submission(s): 946 Problem Description Your input is a series of rectangles, one per line. Each rectangle is sp…
http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=2610 Boring Counting Time Limit: 3000ms   Memory limit: 65536K  有疑问?点这里^_^ 题目描述     In this problem you are given a number sequence P consisting of N integer and Pi is the ith ele…