原题题面看不懂的可以看下面的\(CJ\)版中文题面 $ $ \(CJ\)版: $ $ 这道题是\(JOI\)的\(T4\),放到联赛大概就是\(Day2,T3\)的难度 $ $ \(5\)分: 这一档分就是一个分类讨论,因为\(N=2\) 一共有\(3\)种情况可以出现最优解: \(1\)带球跑到\(2\),\(ans=C*(\left|h1-h2\right|+\left|w1-w2\right|)\) \(1\)带球纵向跑到与\(2\)齐平,踢球,\(ans=C*\left|h1-h2\ri…

 Problem A: Football (aka Soccer)  The Problem Football the most popular sport in the world (americans insist to call it "Soccer", but we will call it "Football"). As everyone knows, Brasil is the country that have most World Cup title…

[JOI2017/2018]美術展 题目大意: 有\(n(n\le5\times10^5)\)个物品,每个物品有两个属性:尺寸\(A_i\)和收益\(B_i\).从中选取一个子集,总收益为\(\sum B_i-\max\{A_i\}-\min\{A_i\}\).求总收益最大值. 思路: 将所有物品按照\(A_i\)排序,\(B_i\)前缀和记作\(S_i\).答案相当于\(\max\{S_i-A_i+A_j-S_{j-1}\}\).维护\(A_j-S_{j-1}\)前缀\(\max\)即可. 源…

Football the most popular sport in the world (americans insist to call it "Soccer", but we will call it "Football"). As everyone knows, Brasil is the country that have most World Cup titles (four of them: 1958, 1962, 1970 and 1994). As…

A 1-1 tie at home was sufficient for Guangzhou Evergrande to clinch the Asian Champions League title, the first Chinese club to win the tournament in more than two decades. clinch:拥吻,钉牢 tournament:锦标赛,联赛 on aggregate:总共,总计 It was difficult to spot a…

Chef and Big Soccer   Problem code: CHEFSOC2 Tweet     ALL SUBMISSIONS All submissions for this problem are available. Read problems statements in Mandarin Chinese, Russian and Vietnamese as well. Chef is a big fan of soccer! He loves soccer so much,…

Description All submissions for this problem are available. Read problems statements in Mandarin Chinese, Russian and Vietnamese as well. Chef is a big fan of soccer! He loves soccer so much, that he even invented soccer for his pet dogs! Here are th…

time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard output The country Treeland consists of n cities connected with n - 1 bidirectional roads in such a way that it's possible to reach every city starting…

题意:给指定数量的数字“1”,“2”,“3”……,“9”.用所有这些数字加上任意个0组成一个数,要求数能被11整除,且数的位数尽量小. 能被11整除的数有一个特点,奇数位数字之和与偶数位之和的差为11的倍数. 所以想到把所有数字分成两部分,即奇数位部分和偶数位部分,两部分的差相0即能被11整除(MOD 11). 所求可以化为,其中一部分%11的余数为所有数字之和%11的余数的一半. dp[k][r] := 能否找到 任意k个数之和 %11 == R #include<bits/stdc++.h>…

题目分析: 好像跑得很快,似乎我是第一个启发式合并的. 把玩具看成区间.首先很显然如果有两个玩具的进出时间有$l1<l2<r1<r2$的关系,那么这两个玩具一定在不同的栈中间. 现在假设一定有解,我们怎么得到答案呢?排序会使得计算变得方便,下面我们按照左端点排序. 想象一条扫描线,从左往右,当它遇到了一个区间的左端点的时候,我们尝试着将原先不在一起的合并,所有和这个不同栈的都被合并. 我们可以想象一个并查集,使用堆维护并查集.堆内存储并查集内元素的右端点.在最外面再用一个大堆来存储每个并…