2019 杭电多校 10 1007 题目链接:HDU 6697 比赛链接:2019 Multi-University Training Contest 10 Problem Description The closest pair of points problem is a well-known problem of computational geometry. In this problem, you are given \(n\) points in the Euclidean plan…

首先最容易想到的就是N2暴力枚举所有线段去找最小值,但是这样会做了许多无用功.我们可以先对线段排序,使得线段最左侧的端点按照x轴y轴排序,然后我们可以限定在这个线段的矩形框内的所有线段才有可能产生最小值,每次查询对于第i条线段的最近距离,如果第j条线段的最左侧点的x与第i条线段的最右侧点的x差值大于ans,那么可以直接break,之后枚举是没有任何意义的,一定会大于ans,所以加了这部分剪枝复杂度就压缩了很大部分. // ——By DD_BOND //#include<bits/stdc++.h…

[本文链接] http://www.cnblogs.com/hellogiser/p/closest-pair-problem.html [题目] 给定平面上N个点的坐标,找出距离最近的两个点之间的距离. [蛮力法] 对于n个点,一共可以组成n(n-1)/2对点对,对这n(n-1)/2对点对逐对进行距离计算,通过循环求得点集中的最近点对.时间复杂度为T(n)=n^2. C++ Code  123456789101112131415161718192021222324252627282930313…

HDU 5877 Weak Pair(弱点对) Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others) Description 题目描述 You are given a rooted tree of N nodes, labeled from 1 to N. To the ith node a non-negative value ai is assigned. An ordere…

C. The Closest Pair Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/312/problem/C Description Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane…

题意,给出n个点的坐标,找出两点间最近的距离,如果小于10000就输出INFINITY. 纯暴力是会超时的,所以得另辟蹊径,用分治算法. 递归思路将点按坐标排序后,分成两块处理,最近的距离不是在两块中的一块中,就会存在于跨越中线的点对中. 查找跨越中线的点比较麻烦,之前已经求出两块中的最小距离,只要在x范围在[m-d,m+d]的点中找对,更新最小距离,最后返回最小距离即可. 代码: /* * Author: illuz <iilluzen[at]gmail.com> * Blog: http:…

Problem JThe Closest Pair ProblemInput: standard inputOutput: standard outputTime Limit: 8 secondsMemory Limit: 32 MB Given a set of points in a two dimensional space, you will have to find the distance between the closest two points. Input The input…

//树形DP+树状数组 HDU 5877 Weak Pair // 思路:用树状数组每次加k/a[i],每个节点ans+=Sum(a[i]) 表示每次加大于等于a[i]的值 // 这道题要离散化 #include <bits/stdc++.h> using namespace std; #define LL long long typedef pair<int,int> pii; const double inf = 123456789012345.0; const LL MOD…

Dylans loves tree Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 1444    Accepted Submission(s): 329 Problem Description Dylans is given a tree with N nodes. All nodes have a value A[i].Nodes…

Multiply game Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3224    Accepted Submission(s): 1173 Problem Description Tired of playing computer games, alpc23 is planning to play a game on numbe…

问题 H: Planar map 时间限制: 1 Sec  内存限制: 128 MB提交: 24  解决: 22[提交][状态][讨论版] 题目描述 Tigher has work for a long time in a famous company.One day she is given a planar map(look at the following) and a point.The planar map can be regarded as a polygon.The planar…

HDU 1394 Minimum Inversion Number(线段树求最小逆序数对) ACM 题目地址:HDU 1394 Minimum Inversion Number 题意:  给一个序列由[1,N]构成.能够通过旋转把第一个移动到最后一个.  问旋转后最小的逆序数对. 分析:  注意,序列是由[1,N]构成的,我们模拟下旋转,总的逆序数对会有规律的变化.  求出初始的逆序数对再循环一遍即可了. 至于求逆序数对,我曾经用归并排序解过这道题:点这里.  只是因为数据范围是5000.所以全…

首先判断是不是凸多边形 然后判断圆是否在凸多边形内 不知道给出的点是顺时针还是逆时针,所以用判断是否在多边形内的模板,不用是否在凸多边形内的模板 POJ 1584 A Round Peg in a Ground Hole(判断凸多边形,点到线段距离,点在多边形内) #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath&g…

Minimum Inversion Number Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 17737    Accepted Submission(s): 10763 Problem Description The inversion number of a given number sequence a1, a2, ...,…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5029 Problem Description The soil is cracking up because of the drought and the rabbit kingdom is facing a serious famine. The RRC(Rabbit Red Cross) organizes the distribution of relief grain in the disa…

Vases and Flowers Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others) Total Submission(s): 3220    Accepted Submission(s): 1273 Problem Description Alice is so popular that she can receive many flowers everyday. She ha…

链接:线段树求矩形面积并 扫描线+离散化 1.给定平面上若干矩形,求出被这些矩形覆盖过至少两次的区域的面积. 2.看完线段树求矩形面积并 的方法后,再看这题,求的是矩形面积交,类同. 求面积时,用被覆盖2次以上的那一段乘以扫描线的距离即可,具体实现见代码. 3. /* HDU 1255 覆盖的面积 求矩形面积交(离散化+线段树) 给定一些矩形 求被这些矩形覆盖过至少两次的区域的面积 这里的方法是:线段树求矩形面积交 扫描线+离散化 左右扫描(x轴扫描),把y轴上的线段离散化 */ #includ…

Segments Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 14178   Accepted: 4521 Description Given n segments in the two dimensional space, write a program, which determines if there exists a line such that after projecting these segments…

Color the ball Time Limit: 9000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 7941    Accepted Submission(s): 4070 Problem Description N个气球排成一排,从左到右依次编号为1,2,3....N.每次给定2个整数a b(a <= b),lele便为骑上他的“小飞鸽"牌电动车从气球…

D. Vika and Segments     Vika has an infinite sheet of squared paper. Initially all squares are white. She introduced a two-dimensional coordinate system on this sheet and drew n black horizontal and vertical segments parallel to the coordinate axes.…

Hyperspace Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 1023    Accepted Submission(s): 492 Problem Description The great Mr.Smith has invented a hyperspace particle generator. The device i…

Just a Hook Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 27359    Accepted Submission(s): 13593 Problem Description In the game of DotA, Pudge’s meat hook is actually the most horrible thing…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1255 Description 给定平面上若干矩形,求出被这些矩形覆盖过至少两次的区域的面积.   Input 输入数据的第一行是一个正整数T(1<=T<=100),代表测试数据的数量.每个测试数据的第一行是一个正整数N(1<=N<=1000),代表矩形的数量,然后是N行数据,每一行包含四个浮点数,代表平面上的一个矩形的左上角坐标和右下角坐标,矩形的上下边和X轴平行,左右边和Y轴平行.坐…

做这道题之前,建议先做POJ 1151  Atlantis,经典的扫描线求矩阵的面积并 参考连接: http://www.cnblogs.com/scau20110726/archive/2013/04/13/3018702.html 线段树辅助——扫描线法计算矩形周长并(轮廓线):http://www.cnblogs.com/scau20110726/archive/2013/04/13/3018687.htmlhttp://blog.csdn.net/ophunter/article/det…

题目链接 题意 : 一个有n段长的金属棍,开始都涂上铜,分段涂成别的,金的值是3,银的值是2,铜的值是1,然后问你最后这n段总共的值是多少. 思路 : 线段树的区间更新.可以理解为线段树成段更新的模板题. //HDU 1698 #include <cstdio> #include <cstring> #include <iostream> using namespace std; ],p[] ; //lz延迟标记,每次更新不需要更新到底,使得更新延迟到下次更新或者查询的…

题目链接:http://acm.timus.ru/problem.aspx?space=1&num=1987 题意:给定n条线段,每两条线段要么满足没有公共部分,要么包含.给出m个询问,求当前点被覆盖的最小长度的线段编号. 由于线段不存在部分相交的情况,因此,直接按照输入顺序覆盖区间就可以了,因为后覆盖的线段更短. //STATUS:C++_AC_187MS_6805KB #include <functional> #include <algorithm> #include…

A Round Peg in a Ground Hole Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 4438   Accepted: 1362 Description The DIY Furniture company specializes in assemble-it-yourself furniture kits. Typically, the pieces of wood are attached to on…

D. Vika and Segments 题目连接: http://www.codeforces.com/contest/610/problem/D Description Vika has an infinite sheet of squared paper. Initially all squares are white. She introduced a two-dimensional coordinate system on this sheet and drew n black hor…

题意: 给你n个数的序列a,q个询问,每个询问给l,r,求在下标i在[l,r]的区间任意两个数的最大公约数中的最大值 分析: 有了hdu3333经验,我们从左向右扫序列,如果当前数的约数在前面出现过,那这个约数可能就是最大的答案.所以我们枚举当前数的所有约数,用线段树维护区间最大值,查询序列离线处理保证查询的正确. #include <map> #include <set> #include <list> #include <cmath> #include…

Wow! Such Sequence! 题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=4893 Description Recently, Doge got a funny birthday present from his new friend, Protein Tiger from St. Beeze College. No, not cactuses. It's a mysterious blackbox. After some resea…