AC日记——[USACO15DEC]最大流Max Flow 洛谷 P3128
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 between pairs of stalls (
). For the
th such pair, you are told two stalls
and
, endpoints of a path along which milk is being pumped at a unit rate. FJ is concerned that some stalls might end up overwhelmed with all the milk being pumped through them, since a stall can serve as a waypoint along many of the
paths along which milk is being pumped. Please help him determine the maximum amount of milk being pumped through any stall. If milk is being pumped along a path from
to
, then it counts as being pumped through the endpoint stalls
and
, as well as through every stall along the path between them.
FJ给他的牛棚的N(2≤N≤50,000)个隔间之间安装了N-1根管道,隔间编号从1到N。所有隔间都被管道连通了。
FJ有K(1≤K≤100,000)条运输牛奶的路线,第i条路线从隔间si运输到隔间ti。一条运输路线会给它的两个端点处的隔间以及中间途径的所有隔间带来一个单位的运输压力,你需要计算压力最大的隔间的压力是多少。
输入输出格式
输入格式:
The first line of the input contains and
.
The next lines each contain two integers
and
(
) describing a pipe
between stalls and
.
The next lines each contain two integers
and
describing the endpoint
stalls of a path through which milk is being pumped.
输出格式:
An integer specifying the maximum amount of milk pumped through any stall in the
barn.
输入输出样例
5 10
3 4
1 5
4 2
5 4
5 4
5 4
3 5
4 3
4 3
1 3
3 5
5 4
1 5
3 4
9 思路:
裸树剖;
(感觉正确的代码样例没过,错误的代码ac。。。) 来,上代码:
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm> #define maxn 50005 using namespace std; struct TreeNodeType {
int l,r,dis,mid,flag;
};
struct TreeNodeType tree[maxn<<]; struct EdgeType {
int to,next;
};
struct EdgeType edge[maxn<<]; int if_z,n,m,cnt,head[maxn],deep[maxn],id[maxn];
int size[maxn],top[maxn],f[maxn]; char Cget; inline void in(int &now)
{
now=,if_z=,Cget=getchar();
while(Cget>''||Cget<'')
{
if(Cget=='-') if_z=-;
Cget=getchar();
}
while(Cget>=''&&Cget<='')
{
now=now*+Cget-'';
Cget=getchar();
}
now*=if_z;
} inline void edge_add(int u,int v)
{
cnt++;
edge[cnt].to=v;
edge[cnt].next=head[u];
head[u]=cnt;
} void search_1(int now,int fa)
{
int pos=cnt++;
f[now]=fa,deep[now]=deep[fa]+;
for(int i=head[now];i;i=edge[i].next)
{
if(edge[i].to==fa) continue;
search_1(edge[i].to,now);
}
size[now]=cnt-pos;
} void search_2(int now,int chain)
{
id[now]=++cnt,top[now]=chain;
int pos=;
for(int i=head[now];i;i=edge[i].next)
{
if(edge[i].to==f[now]) continue;
if(size[edge[i].to]>size[pos]) pos=edge[i].to;
}
if(pos==) return ;
search_2(pos,chain);
for(int i=head[now];i;i=edge[i].next)
{
if(edge[i].to==f[now]||edge[i].to==pos) continue;
search_2(edge[i].to,edge[i].to);
}
} void tree_build(int now,int l,int r)
{
tree[now].l=l,tree[now].r=r;
if(l==r) return ;
tree[now].mid=(l+r)>>;
tree_build(now<<,l,tree[now].mid);
tree_build(now<<|,tree[now].mid+,r);
} void tree_change(int now,int l,int r)
{
if(tree[now].l==l&&tree[now].r==r)
{
tree[now].dis++;
tree[now].flag++;
return ;
}
if(tree[now].flag)
{
tree[now<<].dis+=tree[now].flag,tree[now<<|].dis+=tree[now].flag;
tree[now<<].flag+=tree[now].flag,tree[now<<|].flag+=tree[now].flag;
tree[now].flag=;
}
if(l>tree[now].mid) tree_change(now<<|,l,r);
else if(r<=tree[now].mid) tree_change(now<<,l,r);
else
{
tree_change(now<<,l,tree[now].mid);
tree_change(now<<|,tree[now].mid+,r);
}
tree[now].dis=max(tree[now<<].dis,tree[now<<|].dis);
} int main()
{
in(n),in(m);int u,v;
for(int i=;i<n;i++)
{
in(u),in(v);
edge_add(u,v);
edge_add(v,u);
}
cnt=,search_1(,);
cnt=,search_2(,);
tree_build(,,n);
while(m--)
{
in(u),in(v);
while(top[u]!=top[v])
{
if(deep[top[u]]<deep[top[v]]) swap(u,v);
tree_change(,id[top[u]],id[u]);
u=f[top[u]];
}
//if(u==v) continue;
if(deep[u]>deep[v]) swap(u,v);
tree_change(,id[u],id[v]);
}
cout<<tree[].dis;
return ;
}
AC日记——[USACO15DEC]最大流Max Flow 洛谷 P3128的更多相关文章
- AC日记——[USACO09JAN]全流Total Flow 洛谷 P2936
题目描述 Farmer John always wants his cows to have enough water and thus has made a map of the N (1 < ...
- 洛谷P3128 [USACO15DEC]最大流Max Flow
P3128 [USACO15DEC]最大流Max Flow 题目描述 Farmer John has installed a new system of N-1N−1 pipes to transpo ...
- P3128 [USACO15DEC]最大流Max Flow(LCA+树上差分)
P3128 [USACO15DEC]最大流Max Flow 题目描述 Farmer John has installed a new system of pipes to transport mil ...
- luoguP3128 [USACO15DEC]最大流Max Flow 题解(树上差分)
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 ...
- 洛谷P3128 [USACO15DEC]最大流Max Flow [树链剖分]
题目描述 Farmer John has installed a new system of pipes to transport milk between the stalls in his b ...
- [USACO15DEC]最大流Max Flow(树上差分)
题目描述: Farmer John has installed a new system of N−1N-1N−1 pipes to transport milk between the NNN st ...
- 洛谷P3128 [USACO15DEC]最大流Max Flow [倍增LCA]
题目描述 Farmer John has installed a new system of pipes to transport milk between the stalls in his b ...
- 洛谷P3128 [USACO15DEC]最大流Max Flow(树上差分)
题意 题目链接 Sol 树上差分模板题 发现自己傻傻的分不清边差分和点差分 边差分就是对边进行操作,我们在\(u, v\)除加上\(val\),同时在\(lca\)处减去\(2 * val\) 点差分 ...
- 洛谷 P3128 [USACO15DEC]最大流Max Flow
题目描述 \(FJ\)给他的牛棚的\(N(2≤N≤50,000)\)个隔间之间安装了\(N-1\)根管道,隔间编号从\(1\)到\(N\).所有隔间都被管道连通了. \(FJ\)有\(K(1≤K≤10 ...
随机推荐
- Python基础——列表(list)
创建列表(list) 通过[]来创建list结构,里面放任何类型都可以,没有长度限制. list1=[] type(list1) list1=[1,2,3,4] list1 list1=['] lis ...
- opencv和numpy的安装
近日,学姐让我们切割图片,查了一下资料,发现我需要安装opencv和numpy.但是在安装过程中却出现了很多小问题,我在此结合自和自己的安装经验和网上查找的资料,做一个笔记. 1.opencv的安装 ...
- python基础-面向对象的三大特征
继承 单继承 父类 基类 子类 派生类 继承:是面向对象软件技术当中的一个概念,如果一个类别A“继承自”另一个类别B,就把这个A称为“B的子类别”,而把B称为“A的父类别”也可以称“B是A的超类”. ...
- linux学习-systemd-journald.service 简介
过去只有 rsyslogd 的年代中,由于 rsyslogd 必须要开机完成并且执行了 rsyslogd 这个 daemon 之 后,登录文件才会开始记录.所以,核心还得要自己产生一个 klogd 的 ...
- while True 死循环
while True 死循环示例: count = 0 #给count设置变量为0 while True: count += 1 #每循环一次,count+1 : count += 1 等同于coun ...
- javaweb通过接口来实现多个文件压缩和下载(包括单文件下载,多文件批量下载)
原博客地址:https://blog.csdn.net/weixin_37766296/article/details/80044000 将多个文件压缩并下载下来:(绿色为修改原博客的位置) 注意:需 ...
- Educational Codeforces Round 2 Edge coloring of bipartite graph
题意: 输入一个二分图,用最少的颜色数给它的每条边染色,使得同一个顶点连的边中颜色互不相同. 输出至少需要的颜色数和任意一种染色方案. 分析: 证明不会,只说一下(偷瞄巨巨代码学到的)做法. 假设点的 ...
- IDEA字体颜色快速导入辅助工具设置
原创链接:https://www.cnblogs.com/ka-bu-qi-nuo/p/9181954.html 程序员开发大多数都是使用IDE进行代码开发的,这样能快速的开发出需要的项目.之前一直 ...
- poj2449 Remmarguts' Date K短路 A*
K短路裸题. #include <algorithm> #include <iostream> #include <cstring> #include <cs ...
- 深入了解ASO
ASO对于一些人来说可能很陌生,很多人都听说过SEO,没有听说过ASO(我也是最近才知道这个领域),因为这是一个数字营销的一个新领域.ASO(App Store Optimization)是为了让自己 ...