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
- DELL PowerEdge R620安装Windows server(你想将windows安装在何处”找不到任何本地磁盘,“找不到驱动器”)已解决!
你可能碰到过DELL服务器上安装Windows server系列系统时无法识别或找不到硬盘的问题,对于DELL PowerEdge11-14代机器的,大家可以采用DELL的Lifecycle cont ...
- C语言中的32个关键字
C语言中的32个关键字 数据类型关键字(12个) (1) char:声明字符型变量或函数 (2) double:声明双精度变量或函数 (3) enum:声明美剧类型 (4) ...
- 【HIHOCODER 1142】 三分·三分求极值
描述 这一次我们就简单一点了,题目在此: 在直角坐标系中有一条抛物线y=ax^2+bx+c和一个点P(x,y),求点P到抛物线的最短距离d. 输入 第1行:5个整数a,b,c,x,y.前三个数构成抛物 ...
- 线段树: CDOJ1598-加帕里公园的friends(区间合并,单点更新)
加帕里公园的friends Time Limit: 3000/1000MS (Java/Others) Memory Limit: 131072/131072KB (Java/Others) 我还有很 ...
- Linux学习-systemctl 针对 timer 的配置文件
如何使用 systemd 内建的 time 来处理各种任务呢? systemd.timer 的优势 在 archlinux 的官网 wiki 上面有提到,为啥要使用 systemd.timer 呢? ...
- HDU 5510 Bazinga KMP
题意: 给\(n(1 \leq n \leq 500)\)个字符串,求一个最大的\(i\),使得存在一个\(S_{j}\)不是\(S_i\)的子串. 分析: 维护两个指针\(l,r\) 那么有两种情况 ...
- 下拉列表 Spinner
在Web开发中,HTML提供了下拉列表的实现,就是使用<select>元素实现一个下拉列表,在其中每个下拉列表项使用<option>表示即可.这是在Web开发中一个必不可少的交 ...
- 《Scrum实战》第1课【知易行难】全团课后任务汇总
1组 孟帅(班长) kecyru 2017-7-5 http://kecyru.blog.163.com/blog/static/27416617320176411513013 htt ...
- php-数据库连接类
<?php class DB{ var $host; var $user; var $pwd; var $dbname; var $conn; function DB($host,$user,$ ...
- oracle整体结构-内存结构、物理结构、逻辑结构、进程
Oracle的体系结构大体上分为两部分:Instance(实例)和Database(数据库). Instance(实例) :在Oracle Instance中主要包含了SGA以及一些进程(例如:PMO ...