4390: [Usaco2015 dec]Max Flow Time Limit: 10 Sec  Memory Limit: 128 MBSubmit: 177  Solved: 113[Submit][Status][Discuss] Description Farmer John has installed a new system of N−1 pipes to transport milk between the N stalls in his barn (2≤N≤50,000), c…

[Usaco2015 dec]Max Flow Time Limit: 10 Sec  Memory Limit: 128 MBSubmit: 353  Solved: 236[Submit][Status][Discuss] Description Farmer John has installed a new system of N−1 pipes to transport milk between the N stalls in his barn (2≤N≤50,000), conveni…

BZOJ4390: [Usaco2015 dec]Max Flow Description Farmer John has installed a new system of N−1 pipes to transport milk between the N stalls in his barn (2≤N≤50,000), conveniently numbered 1…N. Each pipe connects a pair of stalls, and all stalls are conn…

题目大意:给出一棵树,n(n<=5w)个节点,k(k<=10w)次修改,每次给定s和t,把s到t的路径上的点权+1,问k次操作后最大点权. 对于每次修改,给s和t的点权+1,给lca(s,t)和lca(s,t)的父亲的点权-1,每一个点的权就是它与它的子树权和,实际上就是树上的差分,又涨姿势了... 代码如下: uses math; type point=^rec; rec=record data:longint; next:point; end; var n,m,x,y,i,ans,fa,k…

题目描述 Farmer John has installed a new system of N−1 pipes to transport milk between the N stalls in his barn (2≤N≤50,000), conveniently numbered 1…N. Each pipe connects a pair of stalls, and all stalls are connected to each-other via paths of pipes.FJ…

Time Limit: 10 Sec  Memory Limit: 128 MBSubmit: 204  Solved: 129[Submit][Status][Discuss] Description Farmer John has installed a new system of N−1 pipes to transport milk between the N stalls in his barn (2≤N≤50,000), conveniently numbered 1…N. Each…

[BZOJ4391][Usaco2015 dec]High Card Low Card(贪心) 题面 BZOJ 题解 预处理前缀后缀的结果,中间找个地方合并就好了. #include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<cmath> #include<algorithm> #include<set> using na…

题意: T,A,B.T是上限.A和B可以随意吃但是不能超过T.有一次将吃的东西/2的机会.然后可以继续吃,不能超过T.问最多可以吃多少. =>我们先处理不能/2可以吃到哪些.然后弄个双指针扫一扫就可以了TAT #include<cstdio> #include<cstring> #include<cctype> #include<algorithm> using namespace std; #define rep(i,s,t) for(int i=s…

洛谷 P3128 [USACO15DEC]最大流Max Flow 洛谷传送门 JDOJ 3027: USACO 2015 Dec Platinum 1.Max Flow JDOJ传送门 Description Farmer John has installed a new system of N−1pipes to transport milk between the ), conveniently numbered. Each pipe connects a pair of stalls, a…

题目描述 Farmer John has installed a new system of  pipes to transport milk between the  stalls in his barn (), conveniently numbered . Each pipe connects a pair of stalls, and all stalls are connected to each-other via paths of pipes. FJ is pumping milk…

First problem to learn Max Flow. Ford-Fulkerson is a group of algorithms - Dinic is one of it.It is an iterative process: we use BFS to check augament-ability, and use DFS to augment it. Here is the code with my comments from its tutorial #include <c…

Max Flow 题目描述 Farmer John has installed a new system of N−1 pipes to transport milk between the N stalls in his barn (2≤N≤50,000), conveniently numbered 1…N. Each pipe connects a pair of stalls, and all stalls are connected to each-other via paths of…

问题描述 给定g个group,n个id,n<=g.我们将为每个group分配一个id(各个group的id不同).但是每个group分配id需要付出不同的代价cost,需要求解最优的id分配方案,使得整体cost之和最小. 例子 例如以下4个group,三个id,value矩阵A: value id1 id2 id3 H1 4 3 0 H2 1 0 0 H3 2 0 2 H4 3 1 0 id_i分配给H_j的代价\(changing cost[i, j]=\sum(A[j,:])-A[j,i]…

[Luogu 3128] USACO15DEC Max Flow 最近跟 LCA 干上了- 树剖好啊,我再也不想写倍增了. 以及似乎成功转成了空格选手 qwq. 对于每两个点 S and T,求一下 LCA 顺便树上差分,最后求差分数组的前缀和并找出最大值输出就行了. (PS:最近考前训练不开 C++11,所以如果看见我写了奇怪的 define 请自动无视QAQ!) #include <algorithm> #include <cstdio> #define nullptr NUL…

4397: [Usaco2015 dec]Breed Counting Time Limit: 10 Sec  Memory Limit: 128 MB Description Farmer John's N cows, conveniently numbered 1…N, are all standing in a row (they seem to do so often that it now takes very little prompting from Farmer John to…

P3128 [USACO15DEC]最大流Max Flow 题目描述 Farmer John has installed a new system of N-1N−1 pipes to transport milk between the NN stalls in his barn (2 \leq N \leq 50,0002≤N≤50,000), conveniently numbered 1 \ldots N1…N. Each pipe connects a pair of stalls,…

P3128 [USACO15DEC]最大流Max Flow 题目描述 Farmer John has installed a new system of  pipes to transport milk between the  stalls in his barn (), conveniently numbered . Each pipe connects a pair of stalls, and all stalls are connected to each-other via path…

4397: [Usaco2015 dec]Breed Counting Time Limit: 10 Sec  Memory Limit: 128 MB Submit: 29  Solved: 25 [Submit][Status][Discuss] Description Farmer John's N cows, conveniently numbered 1-N, are all standing in a row (they seem to do so often that it now…

链接一下题目:luoguP3128 [USACO15DEC]最大流Max Flow(树上差分板子题) 如果没有学过树上差分,抠这里(其实很简单的,真的):树上差分总结 学了树上差分,这道题就极其显然了,不就是把每一条运输路线差分进去,那就是板子了啊. 树上差分还是很有用的,比较容易写,这种询问很少的题目去敲那么长(还容易出玄学错误)的树剖很浪费,用树上差分就很快了!(//...微笑...\\) 上一波代码: #include<iostream> #include<cstdlib>…

bzoj4397[Usaco2015 dec]Breed Counting 题意: 给定一个长度为N的序列,每个位置上的数只可能是1,2,3中的一种.有Q次询问,每次给定两个数a,b,请分别输出区间[a,b]里数字1,2,3的个数.n≤100000,q≤100000. 题解: 裸前缀和. 代码: #include <cstdio> #include <cstring> #include <algorithm> #define inc(i,j,k) for(int i=j…

bzoj4396[Usaco2015 dec]High Card Wins 题意: 一共有2n张牌,Alice有n张,Bob有n张,每一局点数大的赢.知道Bob的出牌顺序,求Alice最多能赢几局.n≤50000. 题解: 贪心.将Alice和Bob的牌按点数大小排序,然后如果Alice当前牌能赢Bob当前牌就ans++否则就不断调整Bob的当前牌直到Alice当前牌能赢Bob当前牌. 代码: #include <cstdio> #include <cstring> #includ…

bzoj4395[Usaco2015 dec]Switching on the Lights 题意: n*n个房间,奶牛初始在(1,1),且只能在亮的房间里活动.每当奶牛经过一个房间,就可以打开这个房间里控制其它房间灯的开关.问奶牛最多可点亮多少个房间.n≤100. 题解: 因为只要一个房间灯亮了,它将一直亮着,所以可以做bfs,每次由队列中的节点扩展可以到的节点.然而这样做不行,因为可能之前尝试过不能到达的房间的灯可以在之后到达的房间里被打开.解决方法是不停做bfs,直到答案不再更新. 代码:…

最大流算法,解决的是从一个起点到一个终点,通过任何条路径能够得到的最大流量. 有个Edmond-Karp算法: 1. BFS找到一条增广路径:算出这条路径的最小流量(所有边的最小值)increase: 2. 然后更新路径上的权重(流量),正向边加上increase,反向边减去increase; 3. 重复1,直到没有增广路径: 可以证明的是在使用最短路增广时增广过程不超过V*E次,每次BFS的时间都是O(E),所以Edmonds-Karp的时间复杂度就是O(V*E^2). 图的BFS和DFS的时…

题目描述 Farmer John has installed a new system of  pipes to transport milk between the  stalls in his barn (), conveniently numbered . Each pipe connects a pair of stalls, and all stalls are connected to each-other via paths of pipes. FJ is pumping milk…

同运输计划. #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #define maxv 50050 #define maxe 100500 using namespace std; struct edge { int v,nxt; }e[maxe]; ],ans=,dis[maxv],g[maxv],nume=; void addedge(int u,int…

题意比较难理解,就是给你n个点的树,然后给你m个修改操作,每一次修改包括一个点对(x, y),意味着将x到y所有的点权值加一,最后问你整个树上的点权最大是多少. 比较裸的树链剖分了,感谢Haild的讲解. 首先第一遍dfs预处理出size,son(重儿子). 第二遍dfs重编号. 然后线段树就可以了. 感觉就是把一棵树弄成一条一条的链,新奇的hash方法. #include <bits/stdc++.h> #define rep(i, a, b) for (int i = a; i <=…

题目描述 Farmer John has installed a new system of  pipes to transport milk between the  stalls in his barn (), conveniently numbered . Each pipe connects a pair of stalls, and all stalls are connected to each-other via paths of pipes. FJ is pumping milk…

题目描述 Farmer John has installed a new system of N-1N−1 pipes to transport milk between the NN stalls in his barn (2 \leq N \leq 50,0002≤N≤50,000), conveniently numbered 1 \ldots N1…N. Each pipe connects a pair of stalls, and all stalls are connected t…

讲解: https://rpdreamer.blog.luogu.org/ci-fen-and-shu-shang-ci-fen #include <bits/stdc++.h> #define read read() #define up(i,l,r) for(register int i = (l);i <= (r);i++) #define down(i,l,r) for(register int i = (l);i >= (r);i--) #define traversal…

正解:贪心+线段树/set库 解题报告: 算辣直接甩链接qwq 恩这题就贪心?从前往后从后往前各推一次然后找一遍哪个地方最大就欧克了,正确性很容易证明 (这里有个,很妙的想法,就是,从后往前推从前往后推可能会有相同的牌嘛,但是其实这个是没有个关系的! 因为,既然有相同的牌那么就必定有多了的牌,然后如果这个多了的牌比重了的牌大我们就放前面,比重了的小我们就放后面,这样就不会影响答案的正确性了…… 哇我觉得这个想法真是太神仙了像我这种菜鸡自己单独想的话是绝对想不到这个的我可能就直接放弃贪心了TT 但…