Watchmen are in a danger and Doctor Manhattan together with his friend Daniel Dreiberg should warn them as soon as possible. There are n watchmen on a plane, the i-th watchman is located at point (xi, yi). They need to arrange a plan, but there are s…

Watchmen are in a danger and Doctor Manhattan together with his friend Daniel Dreiberg should warn them as soon as possible. There are n watchmen on a plane, the i-th watchman is located at point (xi, yi). They need to arrange a plan, but there are s…

Watchmen are in a danger and Doctor Manhattan together with his friend Daniel Dreiberg should warn them as soon as possible. There are n watchmen on a plane, the i-th watchman is located at point (xi, yi). They need to arrange a plan, but there are s…

题意: 求欧几里得距离与曼哈顿距离相等的组数. 分析: 化简后得到xi=xj||yi=yj,即为求x相等 + y相等 - x与y均相等. 代码: #include<iostream> #include<algorithm> using namespace std; const int maxn = 1000000 + 5; typedef pair<long long , long long>pii; pii p[maxn]; bool cmp(pii a, pii b…

Watchmen 直接上中文 Descriptions: 钟表匠们的好基友马医生和蛋蛋现在要执行拯救表匠们的任务.在平面内一共有n个表匠,第i个表匠的位置为(xi, yi). 他们需要安排一个任务计划,但是确发现了一些问题很难解决.马医生从第i个表匠到第j个表匠所需要的时间为|xi - xj| + |yi - yj|.然而蛋蛋所需要的时间为 要想成功完成任务,必须保证两人从第i个表匠出发,同时到达第j个表匠.现在请你计算最多有多少组表匠的位置满足条件 Input 第一行只有一个数n( 1 ≤ n…

题目链接:http://codeforces.com/contest/651/problem/C 思路:结果就是计算同一横坐标.纵坐标上有多少点,再减去可能重复的数量(用map,pair存一下就OK了). #include<bits/stdc++.h> using namespace std; typedef long long ll; typedef pair <int,int> pii; const int N = 2e5 + 5; map <int,int> x,…

传送门 time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output Watchmen are in a danger and Doctor Manhattan together with his friend Daniel Dreiberg should warn them as soon as possible. There are n…

http://codeforces.com/contest/650/problem/A 一开始想了很久都没有考虑到重复点的影响,解欧拉距离和曼哈顿距离相等可以得到 $x_i=x_j$ 或 $y_i=y_j$ ,假如无重复的话就是分别记录 $x$ 和 $y$ 的值然后 $ans+=mx[i]-1,ans+=my[i]-1$ ,就可以了. 那么有重复的话,有两种解决方法,第一种是分类讨论,就是分重复点和同 $x$ 其他点连.重复点和同 $y$ 其他点连.重复点自己连. #include<bits/s…

Description Watchmen are in a danger and Doctor Manhattan together with his friend Daniel Dreiberg should warn them as soon as possible. There are n watchmen on a plane, the i-th watchman is located at point (xi, yi).They need to arrange a plan, but…

题意:两点(x1,y1), (x2,y2)的曼哈顿距离=欧几里得距离 也就是:x1=x2或y1=y2,再删除重合点造成的重复计数即可. #include <stdio.h> #include <string.h> #include <iostream> #include <algorithm> #include <vector> #include <queue> #include <set> #include <ma…

Watchmen CodeForces - 650A Watchmen are in a danger and Doctor Manhattan together with his friend Daniel Dreiberg should warn them as soon as possible. There are n watchmen on a plane, the i-th watchman is located at point (xi, yi). They need to arra…

C. Watchmen 题目连接: http://www.codeforces.com/contest/651/problem/C Description Watchmen are in a danger and Doctor Manhattan together with his friend Daniel Dreiberg should warn them as soon as possible. There are n watchmen on a plane, the i-th watch…

A. Watchmen time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output Watchmen are in a danger and Doctor Manhattan together with his friend Daniel Dreiberg should warn them as soon as possible. The…

Watchmen 题意:有n (1 ≤ n ≤ 200 000) 个点,问有多少个点的开平方距离与横纵坐标的绝对值之差的和相等: 即 = |xi - xj| + |yi - yj|.(|xi|, |yi| ≤ 109) 思路:开始想的是容斥原理,即按x,y分别排序,先计算同x的点,然后在计算同y的点,这时由于相同的点之间的连边已经算过了,这样就不能再算.并且同一个y的点中可以每个点有多个点,算是不好编码的(反正我敲了很久..WA了) 反思:上面的容斥原理是从总体的思路来考虑的,这道题的难点也就是…

C. Watchmen time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output Watchmen are in a danger and Doctor Manhattan together with his friend Daniel Dreiberg should warn them as soon as possible. The…

C. Watchmen time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output Watchmen are in a danger and Doctor Manhattan together with his friend Daniel Dreiberg should warn them as soon as possible. The…

A. Watchmen time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output Watchmen are in a danger and Doctor Manhattan together with his friend Daniel Dreiberg should warn them as soon as possible. The…

A. Joysticks time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Friends are going to play console. They have two joysticks and only one charger for them. Initially first joystick is charged at…

题意:给你\(n\)个点,求这\(n\)个点中,曼哈顿距离和欧几里得距离相等的点对数. 题解: 不难发现,当两个点的曼哈顿距离等于欧几里得距离的时候它们的横坐标或者纵坐标至少有一个相同,可以在纸上画一画,当两点不满足上文所说的情况时,他们的曼哈顿距离一定大于直线距离,因为三角形的两边必定大于第三边,然后我们边输入边求,用桶分别存横和纵坐标,每次出现都将目前桶的值贡献给答案,但是两个相同的点应该只算一次贡献,我们要再开一个桶来减去重复的次数. 代码: int n; map<ll,ll> mpx,…

 cf之路,1,Codeforces Round #345 (Div. 2) ps:昨天第一次参加cf比赛,比赛之前为了熟悉下cf比赛题目的难度.所以做了round#345连试试水的深浅.....       其实这个应该是昨天就写完的,不过没时间了,就留到了今天.. 地址:http://codeforces.com/contest/651/problem/A A. Joysticks time limit per test 1 second memory limit per test 256…

E. Bindian Signalizing Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/problemset/problem/5/E Description Everyone knows that long ago on the territory of present-day Berland there lived Bindian tribes. Their capital was surrounded…

Codeforces #345 Div.1 打CF有助于提高做题的正确率. Watchmen 题目描述:求欧拉距离等于曼哈顿距离的点对个数. solution 签到题,其实就是求有多少对点在同一行或同一列. 时间复杂度:\(O(nlogn)\) Image Preview 题目描述:给定看一张照片的时间,翻页的时间,把图片翻转的时间.一开始屏幕显示第一张照片,可以向左或向右翻,不能跳过还没有看过的图片,方向不对的图片要先翻转再看,看过的不消耗翻转时间与看照片时间,问在一定时间内,最多能看多少张照…

A. Joysticks time limit per test:1 second memory limit per test:256 megabytes input:standard input output:standard output Friends are going to play console. They have two joysticks and only one charger for them. Initially first joystick is charged at…

最近在使用codeblock,所以就先刷一些水题上上手 使用codeblock遇到的问题 1.无法进行编译-------从setting中的编译器设置中配置编译器 2.建立cpp后无法调试------只建立源文件无法调试,需要建立一个工程后才能调试 3.设置断点后,调试不会停止------开启-g模式且工程要建立在一个没有中文名的文件夹下 4.调试中如何查看变量------打开debug中的watch,右键编辑界面的变量可以选择添加变量 水题来源---codeforces(hzwer神犇刷的水题…

上一次我们拿学校的URP做了个小小的demo.... 其实我们还可以把每个学生的证件照爬下来做成一个证件照校花校草评比 另外也可以写一个物理实验自动选课... 但是出于多种原因,,还是绕开这些敏感话题.. 今天,我们来扒一下cf的题面! PS:本代码不是我原创 1. 必要的分析 1.1 页面的获取 一般情况CF的每一个 contest 是这样的: 对应的URL是:http://codeforces.com/contest/xxx 还有一个Complete problemset页面,它是这样的:…

http://codeforces.com/contest/738/problem/D Galya is playing one-dimensional Sea Battle on a 1 × n grid. In this game a ships are placed on the grid. Each of the ships consists of bconsecutive cells. No cell can be part of two ships, however, the shi…

http://codeforces.com/contest/738/problem/C Vasya is currently at a car rental service, and he wants to reach cinema. The film he has bought a ticket for starts in t minutes. There is a straight road of length s from the service to the cinema. Let's…

http://codeforces.com/contest/738/problem/A Polycarp has interviewed Oleg and has written the interview down without punctuation marks and spaces to save time. Thus, the interview is now a string s consisting of n lowercase English letters. There is…

http://codeforces.com/problemset/problem/662/A 题目大意: 给定n(n <= 500000)张卡片,每张卡片的两个面都写有数字,每个面都有0.5的概率是在正面,各个卡牌独立.求把所有卡牌来玩Nim游戏,先手必胜的概率. (⊙o⊙)-由于本人只会在word文档里写公式,所以本博客是图片格式的. Code #include <cstdio> #include <cstring> #include <algorithm> u…

http://codeforces.com/problemset/problem/274/B 题目大意: 给定你一颗树,每个点上有权值. 现在你每次取出这颗树的一颗子树(即点集和边集均是原图的子集的连通图)且这颗树中必须包含节点1 然后将这颗子树中的所有点的点权+1或-1 求把所有点权全部变为0的最小次数(n<=10^5) 题解: 因为每一次的子树中都必须有1,所以我们得知每一次变换的话1的权值都会变化 所以我们以1为根 现在,我们发现,如果一个节点的权值发生变化,那么他的父节点的权值一定发生变…