The Meeting Place Cannot Be Changed Problem Description The main road in Bytecity is a straight line from south to north. Conveniently, there are coordinates measured in meters from the southernmost building in north direction. At some points on the…

B. The Meeting Place Cannot Be Changed time limit per test 5 seconds memory limit per test 256 megabytes input standard input output standard output The main road in Bytecity is a straight line from south to north. Conveniently, there are coordinates…

The Meeting Place Cannot Be Changed 我发现我最近越来越zz了,md 连调程序都不会了,首先要有想法,之后输出如果和期望的不一样就从输入开始一步一步地调啊,tmd现在输出不一样就知道在那等着,有毛用啊. [题目链接]The Meeting Place Cannot Be Changed [题目类型]二分答案 &题解: 二分时间,判断函数比较难想,要先假设所有人都向左走,找到坐标最大的那个,之后在假设所有人都向右走,找到坐标最小的那个,如果最大的<=最小的就行…

地址:http://codeforces.com/contest/782/problem/B 题目: B. The Meeting Place Cannot Be Changed time limit per test 5 seconds memory limit per test 256 megabytes input standard input output standard output The main road in Bytecity is a straight line from…

780B - The Meeting Place Cannot Be Changed 思路: 二分答案: 代码: #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; #define eps 1e-7 #define INF 1e18 #define maxn 60005 int n; double xi[maxn]…

题目链接 The Meeting Place Cannot Be Changed 二分答案即可. check的时候先算出每个点可到达的范围的区间,然后求并集.判断一下是否满足l <= r就好了. eps我设了1e-7. #include <bits/stdc++.h> using namespace std; #define rep(i, a, b) for(int i(a); i <= (b); ++i) + ; ; double mi, ma; struct node{ dou…

                                                               B. The Meeting Place Cannot Be Changed The main road in Bytecity is a straight line from south to north. Conveniently, there are coordinates measured in meters from the southernmost build…

https://vjudge.net/problem/CodeForces-782B B. The Meeting Place Cannot Be Changed time limit per test 5 seconds memory limit per test 256 megabytes input standard input output standard output The main road in Bytecity is a straight line from south to…

B. The Meeting Place Cannot Be Changed time limit per test 5 seconds memory limit per test 256 megabytes input standard input output standard output The main road in Bytecity is a straight line from south to north. Conveniently, there are coordinates…

time limit per test 5 seconds memory limit per test 256 megabytes input standard input output standard output The main road in Bytecity is a straight line from south to north. Conveniently, there are coordinates measured in meters from the southernmo…

Link 题意:给出$n$个坐标$x_i$,$n$个速度$v_i$问使他们相遇的最短时间是多少. 思路:首先可肯定最终相遇位置必定在区间$[0,max(x_i)]$中,二分最终位置,判断左右部分各自所花时间,缩小范围即可. /** @Date : 2017-05-09 22:07:43 * @FileName: 782B.cpp * @Platform: Windows * @Author : Lweleth (SoundEarlf@gmail.com) * @Link : https://gi…

三分显然,要注意EPS必须设成1e-6,设得再小一点都会TLE……坑炸了 #include<cstdio> #include<algorithm> #include<cmath> using namespace std; #define EPS 0.000001 int n,x[60010],v[60010]; double calc(double p) { double res=0; for(int i=1;i<=n;++i) res=max(res,fabs(…

题意:题意:给出n个人的在x轴的位置和最大速度,求n个人相遇的最短时间. 析:二分时间,然后求并集,注意精度,不然会超时. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream>…

题意: The main road in Bytecity is a straight line from south to north. Conveniently, there are coordinates measured in meters from the southernmost building in north direction. At some points on the road there are n friends, and i-th of them is standi…

链接:http://codeforces.com/problemset/problem/782/B 题意: N个点,需要找到一个点使得每个点到这个点耗时最小,每个点都同时开始,且都拥有自己的速度 题解: 对于一个确定的位置,如果耗时最久的点在右边,则这个位置可以往右靠,否则就往左靠,这样,一个二分的解法就形成了 import java.lang.Math; import java.util.Scanner; public class CodeForces_403_B { private stat…

[题目链接]:http://codeforces.com/contest/782/problem/B [题意] 每个人都有一个速度,只能往上走或往下走; 然后让你找一个地方,所有人都能够在t时间内到达; 让t最小. [题解] 很明显的二分了; 二分时间t; 对于某个时间t,这个人都有一个能够到达的唯一区间 [x[i]-v*t..x[i]+v*t] 如果所有人的区间都有交集. 那么就能在那些交集里面随便选一个点了; 区间的交集可以用 最大的左端点小于等于最小的右端点这个依据来判断是否存在 想让答案…

http://codeforces.com/contest/782/problem/B 题意:有n个人,每个人有一个位置和速度,现在要让这n个人都走到同一个位置,问最少需要的时间是多少. 思路:看上去很像二分搜索啊!枚举距离,判断是否有更少的时间,然后发现时间不随着距离单调增减,想起前两天被三分虐了一道题,那就是三分吧. 三分枚举距离,然后判断是否有更少的时间就可以了.比赛时候用了max函数还有精度开太大,超时了,索性直接循环100次. #include <bits/stdc++.h> usi…

题意:给你一张有向图,某人会任意选择起点然后走无穷多步,问是否存在一个点(要求输出)不管他起点在何处怎么走都必经?n<=100005,m<=500005. 标程: #include<bits/stdc++.h> using namespace std; ; int cnt,vis[N],n,m,u,v,a[N],tried[N],head[N]; ]; void add(int x,int y) {num[++cnt].to=y;num[cnt].next=head[x];head…

看来快掉到灰名的蒟蒻涨rating也快... A题模拟一下就好(一开始还sb,, #include<bits/stdc++.h> #define LL long long using namespace std; ]; ],n; int main() { scanf("%d",&n); ; i<=*n; i++) { int x; scanf("%d",&x); ; else tot--; ans=max(ans,tot); } c…

题意:在一维坐标轴上,给定n个点的坐标以及他们的最大移动速度,问他们能聚到某一点处的最短时间. 分析: 1.二分枚举最短时间即可. 2.通过检查当前时间下,各点的最大移动范围之间是否有交集,不断缩小搜索范围. 3.相当于二分枚举找右临界线,符合要求的点都在右边. 4.通过给二分一个查找次数的上界,eg:k=200,而不是dcmp(l, r)<=0,后者会tle. 5.检查是否有交集,就是以第一个点的最大移动区间为标准,不断与其他区间取交集并更新再继续比较. #include<cstdio>…

Codeforces Round #403 (Div. 2, based on Technocup 2017 Finals) 说一点东西: 昨天晚上$9:05$开始太不好了,我在学校学校$9:40$放学我呆到十点然后还要跑回家耽误时间....要不然$D$题就写完了 周末一些成绩好的同学单独在艺术楼上课然后晚上下第一节晚自习和他们在回廊里玩开灯之后再关上一片漆黑真好玩 A.Andryusha and Socks 日常煞笔提.....我竟然$WA$了一次忘了$n<<1$ #include <…

Div1单场我从来就没上过分,这场又剧毒,半天才打出B,C挂了好几次最后还FST了,回紫了. AC:AB Rank:340 Rating:2204-71->2133 Div2.B.The Meeting Place Cannot Be Changed 题目大意:n个人,第i个人位于xi,速度为vi,找到一个点使得所有人到这个点的耗时最小,输出耗时.(n<=60000) 思路:二分答案,知道耗时后可以求出每个人能到达的区间,如果所有区间有交则合法,复杂度O(nlog). #include<…

The Meeting Place Cannot Be Changed CodeForces - 982F Petr is a detective in Braginsk. Somebody stole a huge amount of money from a bank and Petr is to catch him. Somebody told Petr that some luxurious car moves along the roads without stopping. Petr…

A. Andryusha and Socks time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Andryusha is an orderly boy and likes to keep things in their place. Today he faced a problem to put his socks in the…

B. The Meeting Place Cannot Be Changed time limit per test 5 seconds memory limit per test 256 megabytes input standard input output standard output The main road in Bytecity is a straight line from south to north. Conveniently, there are coordinates…

精神不佳,选择了在场外同步划水 没想到实际做起来手感还好,早知道就报名了…… 该打 未完待续233 A. Andryusha and Socks 模拟,模拟大法好.注意每次是先判断完能不能收进柜子,再算桌子上的袜子数量的. /*by SilverN*/ #include<iostream> #include<algorithm> #include<cstring> #include<cstdio> #include<cmath> using na…

B. The Meeting Place Cannot Be Changed time limit per test 5 seconds memory limit per test 256 megabytes input standard input output standard output The main road in Bytecity is a straight line from south to north. Conveniently, there are coordinates…

Given an array of meeting time intervals consisting of start and end times [[s1,e1],[s2,e2],...] (si < ei), find the minimum number of conference rooms required. For example,Given [[0, 30],[5, 10],[15, 20]],return 2. 这道题是之前那道Meeting Rooms的拓展,那道题只让我们是…

Given an array of meeting time intervals consisting of start and end times [[s1,e1],[s2,e2],...] (si < ei), determine if a person could attend all meetings. For example,Given [[0, 30],[5, 10],[15, 20]],return false. 这道题给了我们一堆会议的时间,问我们能不能同时参见所有的会议,这实际…

Meeting time: 2015.November.11th 1:00~2:00 Chairperson:  Serg Melikyan, PTL from Mirantis Meeting summary: 1.Image and package visibility concerns Desc: The general idea is to make sure that if a murano package is public, It should be really availabl…