Bandit in a City 二分+卡常 题目大意: 给定一个 n n n 个节点的树,树的根节点一定是 1 1 1 ,每个节点都有权值 a i a_i ai,每个节点的权值都要往下分配转移,直到叶子节点,问叶子节点中最大的权值最小是多少. 权值可以任意整数分配,但一定要分完。 思路: 最大最小问题,直接考虑二分,二分节点...
}e[maxn*2];voidadd(intu,intv){ e[++cnt].to = v; e[cnt].next = h[u]; h[u] = cnt; } ll lev[maxn]; ll sum[maxn];voiddfs(intu,intpar){ sum[u] = a[u]; lev[u] =0;intcnt =0;for(inti=h[u];i!=-1;i=e[i].next){intv = e[i].to;if(v == par)continue...
CF1436D Bandit in a City 题解 思路 一道CF题 看题目很明显可以使用二分,但又想不到很好的judging函数 所以考虑动态规划 先考虑假如完全理想均分 那么每个叶子节点可以被分到⌈size[u]leaf[u]⌉⌈size[u]leaf[u]⌉ (为什么是上取整??仔细想想 那么现在我们能进行分配的只有当前节点权值 如果每个子节...
题解CF1436D Bandit in a City \(1\)为根的\(n\)个节点的一棵树,每个节点上有\(a_i\)个人,每个人可以选择往任意子节点走,直到走到叶子节点为止,问最后人最多的叶子节点最少有多少人。 最少的最多嘛!那就二分!很容易想到验证的方法,因为一个点可以到下面任意一个叶子,所以记录一下下面叶子最多能容...
What? We thought you died or something. You haven't been around in so long... we divvied up your cut already. Ah, sorry 'bout that. We'll start holding your cut for you again. But you got to come see us at least once a month. Or we'll assume your dead. B
but we're all tired of bosses. But, we liked Split Jack's plan. So if you can get that bitch at Rivet City to buy into our "protection services" we'll split the take with ya. So we don't need ya to be our leader. But we will cut you in on the protection racket. IF you ...
Sometimes it’s competition, sometimes it’s city officials, sometimes it’s opposing politicians, sometimes it’s people who think the back of your sign would make good material for their garage-sale sign. In all cases if you make it harder for your sign to be removed the longer it will...
You'll be asked to pay the following charges by the property at check-in or check-out. Fees may include applicable taxes: A taxis imposed by the city: EUR 1.00 per person, per night Pets Service animals exempt from fees Pets are allowed for an extra charge of EUR 10 per pet, per ...
After my wife lost her entire month's wages to the one-arm bandits in Atlantic City, we've decided to stay away from casinos altogether. See also: bandit one-armed bandit A slot machine designed for gambling in which you pull down a lever on one side that generates a random combination...
Here is a rider, though – Jim is aware that the brand’s in-app navigation is poor (he ran a survey with active users to arrive at the conclusion) and visitors find it difficult to locate the product. To improve navigation, he decides to do an experiment where he creates a variation...