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…

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…

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…

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…

650A Watchmen 637A Voting for Photos 点击查看原题 650A又是一个排序去重的问题,一定要注意数据范围用long long ,而且在写计算组合函数的时候注意也要用long long 虽然10^9没有超过long的范围,但是在计算n*(n-1)/2的过程中超了,所以需要用long long ,否则会出错. #include<iostream> #include<cmath> #include<cstdlib> #include<c…

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 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 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…