Description: 给定一棵树,每次删去叶子,有m个限制,分别为(a,b)表示a需要比b先删,为每个点能否成为最后被删的点 Hint: \(n,m \le 10^5\) Solution: 手模后会发现一个十分不显然的规律: 若a比b先删,则a在以b为根的子树中的点,都不能最后删, 于是这样就转化为这个题了:https://www.cnblogs.com/list1/p/10497877.html 每次直接打个差分标记 这题还要判无解的情况(这谁想得到啊) #include <map>…

首先考虑怎么check一个点是否能被最后一个删除. 可以这么建图,以这个点建有根树,边全部向上指,再加上剩下的有向边. 很明显,这里的一条边的定义就变成了只有删去这个点,才可以删去它指向的点. 因此,只需要建n次图暴力判断是否有环即可. 这样做是n^2的. 考虑加入一条边后,会产生什么影响. 发现这条边会导致一些点答案直接被钦定为零(这些点满足以它们为根一定会存在环) 设边为u---->v 具体分析一下,这些点是所有以v为根建有根树后,u子树内的所有点. 这个可以通过类似换根的分类讨论的方法来找…

首先可以思考一下每次能删去的点有什么性质. 不难发现,每次能删去的点都是入度恰好为 \(1\) 的那些点(包括 \(a_i \rightarrow b_i\) 的有向边). 换句话说,每次能删去的点既要是树上的叶子节点,并且不会被任意一条有向边 \(a_i \rightarrow b_i\) 指向.那么再来思考一下每个点能否走后离开. 因为 \(i\) 号点只能最后离开,那么我们将 \(i\) 看作这课树的根(因为每次只能走叶子).你会发现如果 \(i\) 不能最后走当前仅当会存在一条路径(走有…

题目描述 Bessie正在计划一年一度的奶牛大集会,来自全国各地的奶牛将来参加这一次集会.当然,她会选择最方便的地点来举办这次集会.每个奶牛居住在 N(1<=N<=100,000) 个农场中的一个,这些农场由N-1条道路连接,并且从任意一个农场都能够到达另外一个农场.道路i连接农场A_i和B_i(1 <= A_i <=N; 1 <= B_i <= N),长度为L_i(1 <= L_i <= 1,000).集会可以在N个农场中的任意一个举行.另外,每个牛棚中居…

题目链接 先把\(1\)作为根求每个子树的\(size\),算出把\(1\)作为集会点的代价,不难发现把集会点移动到\(u\)的儿子\(v\)上后的代价为原代价-\(v\)的\(size\)*边权+(总的\(size\)-\(v\)的\(size\))*边权 #include<iostream> #include<cstring> #include<cstdio> #define int long long using namespace std; const int…

题目描述 Bessie is planning the annual Great Cow Gathering for cows all across the country and, of course, she would like to choose the most convenient location for the gathering to take place. Bessie正在计划一年一度的奶牛大集会,来自全国各地的奶牛将来参加这一次集会.当然,她会选择最方便的地点来举办这次集会…

题目描述 Bessie is planning the annual Great Cow Gathering for cows all across the country and, of course, she would like to choose the most convenient location for the gathering to take place. Each cow lives in one of N (1 <= N <= 100,000) different ba…

P2986 [USACO10MAR]伟大的奶牛聚集Great Cow Gat… 题目描述 Bessie is planning the annual Great Cow Gathering for cows all across the country and, of course, she would like to choose the most convenient location for the gathering to take place. Each cow lives in on…

题目描述 Bessie is planning the annual Great Cow Gathering for cows all across the country and, of course, she would like to choose the most convenient location for the gathering to take place. Bessie正在计划一年一度的奶牛大集会,来自全国各地的奶牛将来参加这一次集会.当然,她会选择最方便的地点来举办这次集会…

题目描述 Bessie is planning the annual Great Cow Gathering for cows all across the country and, of course, she would like to choose the most convenient location for the gathering to take place. Bessie正在计划一年一度的奶牛大集会,来自全国各地的奶牛将来参加这一次集会.当然,她会选择最方便的地点来举办这次集会…

题面 题目描述 Bessie is planning the annual Great Cow Gathering for cows all across the country and, of course, she would like to choose the most convenient location for the gathering to take place. Each cow lives in one of N (1 <= N <= 100,000) different…

t1 Convention 题目大意 每一头牛都有一个来的时间,一共有\(n\)辆车,求出等待时间最长的那头牛等待的最小时间. 解法 第一眼看到这道题还以为是\(2018noip\)普及组的t3魔鬼题,但是不一样. 我们因为要查找最大最小值,很容易就想到用二分查找. 那么直接查找答案,也就是等待的时间,然后用\(O(n)\)的复杂度的贪心验证就可以了,放的下且能放进去,那么就放. 总体复杂度是\(O(nlogn)\) \(ps.\)要注意边界的判断. t2 Convention II 题目大意…

题目描述 Bessie is planning the annual Great Cow Gathering for cows all across the country and, of course, she would like to choose the most convenient location for the gathering to take place. Bessie正在计划一年一度的奶牛大集会,来自全国各地的奶牛将来参加这一次集会.当然,她会选择最方便的地点来举办这次集会…

题目描述 Bessie is planning the annual Great Cow Gathering for cows all across the country and, of course, she would like to choose the most convenient location for the gathering to take place. Bessie正在计划一年一度的奶牛大集会,来自全国各地的奶牛将来参加这一次集会.当然,她会选择最方便的地点来举办这次集会…

就按(博主认为的)难度顺序排吧. Sort It Out 分析 容易发现选出的集合一定是所有逆序对的一个最小覆盖集,那么剩下的就一定是一个LIS.仔细想想还可以发现字典序第\(k\)小的最小覆盖集的补集一定是字典序第\(k\)大的LIS,所以找到这个序列字典序第\(k\)大的LIS就好了. 代码 #include <bits/stdc++.h> #define rin(i,a,b) for(register int i=(a);i<=(b);++i) #define irin(i,a,b…

题目描述 Bessie is planning the annual Great Cow Gathering for cows all across the country and, of course, she would like to choose the most convenient location for the gathering to take place. Bessie正在计划一年一度的奶牛大集会,来自全国各地的奶牛将来参加这一次集会.当然,她会选择最方便的地点来举办这次集会…

前置芝士:Here 本文是基于 OI wiki 上的文章加以修改完成,感谢社区的转载支持和其他方面的支持 树形 DP,即在树上进行的 DP.由于树固有的递归性质,树形 DP 一般都是递归进行的. 基础 以下面这道题为例,介绍一下树形 DP 的一般过程. 例题 洛谷 P1352 没有上司的舞会 题目描述 某大学有 $n$ 个职员,编号为 $1 \sim N$.他们之间有从属关系,也就是说他们的关系就像一棵以校长为根的树,父结点就是子结点的直接上司.现在有个周年庆宴会,宴会每邀请来一个职员都会增加一…

传送门 Catch That Cow Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 80273   Accepted: 25290 Description Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a point N (0 ≤ N ≤ 10…

SB贪心,一开始还想着用二分,看了眼黄学长的blog,发现自己SB了... 最小道路=已选取的奶牛/道路总数. #include <iostream> #include <cstdio> #include <algorithm> using namespace std; ]; int n,m,d,l,ans; int main() { scanf("%d%d%d%d",&n,&m,&d,&l); ;i<=n;i+…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2838 Cow Sorting Problem Description Sherlock's N (1 ≤ N ≤ 100,000) cows are lined up to be milked in the evening. Each cow has a unique "grumpiness" level in the range 1...100,000. Since grumpy cow…

[BZOJ1604][Usaco2008 Open]Cow Neighborhoods 奶牛的邻居 试题描述 了解奶牛们的人都知道,奶牛喜欢成群结队．观察约翰的N(1≤N≤100000)只奶牛,你会发现她们已经结成了几个"群"．每只奶牛在吃草的时候有一个独一无二的位置坐标Xi,Yi(l≤Xi,Yi≤[1．.10^9]:Xi,Yi∈整数．当满足下列两个条件之一,两只奶牛i和j是属于同一个群的:   1．两只奶牛的曼哈顿距离不超过C(1≤C≤10^9),即lXi - xil+IYi - Y…

细读cow.osg 转自:http://www.cnblogs.com/mumuliang/archive/2010/06/03/1873543.html 对,就是那只著名的奶牛. //Group节点,可有子节点.Group { UniqueID Group_0         //Gourp名称DataVariance STATIC   //不知道用来干嘛,一般都是staticcullingActive TRUE      //参与culling?num_children 1         …

Cow Bowling Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 13016   Accepted: 8598 Description The cows don't use actual bowling balls when they go bowling. They each take a number (in the range 0..99), though, and line up in a standard…

在linux下,虚拟机的选择方式有很多,比如vmware for linux,virtual box,还有qemu,在以前,使用qemu的人不多,主要是使用起来有些麻烦,但现在随着Openstack的兴起,qemu也得到了很大的发展,现在在Fedora下使用qemu+kvm性能还是很好的,如果再加上spice,就更不错了.但今天还是主要讲讲qemu下使用的几种镜像格式吧! 1. raw raw格式是最简单,什么都没有,所以叫raw格式.连头文件都没有,就是一个直接给虚拟机进行读写的文件.raw不…

Cow Marathon Time Limit: 2000MS   Memory Limit: 30000K Total Submissions: 3195   Accepted: 1596 Case Time Limit: 1000MS Description After hearing about the epidemic of obesity in the USA, Farmer John wants his cows to get more exercise, so he has com…

http://poj.org/problem?id=2184   Description "Fat and docile, big and dumb, they look so stupid, they aren't much fun..." - Cows with Guns by Dana Lyons The cows want to prove to the public that they are both smart and fun. In order to do this,…

Silver Cow Party Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 12674   Accepted: 5651 Description One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1..N is going to attend the big cow party to be held at farm #X (1 ≤ X …

Catch That Cow Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 77420   Accepted: 24457 Description Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a point N (0 ≤ N ≤ 100,00…

Cow Exhibition Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 12420   Accepted: 4964 Description "Fat and docile, big and dumb, they look so stupid, they aren't much fun..." - Cows with Guns by Dana Lyons The cows want to prove to…

Catch That Cow Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 73973   Accepted: 23308 Description Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a point N (0 ≤ N ≤ 100,00…