AC日记——[USACO11DEC]牧草种植Grass Planting 洛谷 P3038
题目描述
Farmer John has N barren pastures (2 <= N <= 100,000) connected by N-1 bidirectional roads, such that there is exactly one path between any two pastures. Bessie, a cow who loves her grazing time, often complains about how there is no grass on the roads between pastures. Farmer John loves Bessie very much, and today he is finally going to plant grass on the roads. He will do so using a procedure consisting of M steps (1 <= M <= 100,000).
At each step one of two things will happen:
FJ will choose two pastures, and plant a patch of grass along each road in between the two pastures, or,
- Bessie will ask about how many patches of grass on a particular road, and Farmer John must answer her question.
Farmer John is a very poor counter -- help him answer Bessie's questions!
给出一棵n个节点的树,有m个操作,操作为将一条路径上的边权加一或询问某条边的权值。
输入输出格式
输入格式:
Line 1: Two space-separated integers N and M
Lines 2..N: Two space-separated integers describing the endpoints of a road.
- Lines N+1..N+M: Line i+1 describes step i. The first character of the line is either P or Q, which describes whether or not FJ is planting grass or simply querying. This is followed by two space-separated integers A_i and B_i (1 <= A_i, B_i <= N) which describe FJ's action or query.
输出格式:
- Lines 1..???: Each line has the answer to a query, appearing in the same order as the queries appear in the input.
输入输出样例
4 6
1 4
2 4
3 4
P 2 3
P 1 3
Q 3 4
P 1 4
Q 2 4
Q 1 4
2
1
2 思路:
裸树剖; 来,上代码:
#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm> #define maxn 100005 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,deep[maxn],f[maxn],size[maxn];
int top[maxn],id[maxn],head[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;
} void search_1(int now,int fa)
{
int pos=cnt++;
deep[now]=deep[fa]+,f[now]=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)
{
int pos=;
top[now]=chain,id[now]=++cnt;
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+=r-l+;
tree[now].flag++;
return ;
}
if(tree[now].flag)
{
tree[now<<].dis+=tree[now].flag*(tree[now].mid-tree[now].l+);
tree[now<<|].dis+=tree[now].flag*(tree[now].r-tree[now].mid);
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=tree[now<<].dis+tree[now<<|].dis;
} int tree_query(int now,int l,int r)
{
if(tree[now].l==l&&tree[now].r==r) return tree[now].dis;
if(tree[now].flag)
{
tree[now<<].dis+=tree[now].flag*(tree[now].mid-tree[now].l+);
tree[now<<|].dis+=tree[now].flag*(tree[now].r-tree[now].mid);
tree[now<<].flag+=tree[now].flag,tree[now<<|].flag+=tree[now].flag;
tree[now].flag=;
}
if(l>tree[now].mid) return tree_query(now<<|,l,r);
else if(r<=tree[now].mid) return tree_query(now<<,l,r);
else return tree_query(now<<,l,tree[now].mid)+tree_query(now<<|,tree[now].mid+,r);
} int main()
{
in(n),in(m);int u,v;
for(int i=;i<n;i++)
{
in(u),in(v);
edge[++cnt].to=v,edge[cnt].next=head[u],head[u]=cnt;
edge[++cnt].to=u,edge[cnt].next=head[v],head[v]=cnt;
}
char type;
cnt=,search_1(,);
cnt=,search_2(,);
tree_build(,,n);
while(m--)
{
cin>>type;in(u),in(v);
if(type=='P')
{
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]);
}
else
{
int pos=;
while(top[u]!=top[v])
{
if(deep[top[u]]<deep[top[v]]) swap(u,v);
pos+=tree_query(,id[top[u]],id[u]);
u=f[top[u]];
}
if(u==v)
{
printf("%d\n",pos);
continue;
}
if(deep[u]>deep[v]) swap(u,v);
printf("%d\n",pos+tree_query(,id[u]+,id[v]));
}
}
return ;
}
AC日记——[USACO11DEC]牧草种植Grass Planting 洛谷 P3038的更多相关文章
- 洛谷P3038 [USACO11DEC]牧草种植Grass Planting
题目描述 Farmer John has N barren pastures (2 <= N <= 100,000) connected by N-1 bidirectional road ...
- 洛谷 P3038 [USACO11DEC]牧草种植Grass Planting
题目描述 Farmer John has N barren pastures (2 <= N <= 100,000) connected by N-1 bidirectional road ...
- 洛谷 P3038 [USACO11DEC]牧草种植Grass Planting(树链剖分)
题解:仍然是无脑树剖,要注意一下边权,然而这种没有初始边权的题目其实和点权也没什么区别了 代码如下: #include<cstdio> #include<vector> #in ...
- P3038 [USACO11DEC]牧草种植Grass Planting
题目描述 Farmer John has N barren pastures (2 <= N <= 100,000) connected by N-1 bidirectional road ...
- [USACO11DEC]牧草种植Grass Planting
图很丑.明显的树链剖分,需要的操作只有区间修改和区间查询.不过这里是边权,我们怎么把它转成点权呢?对于E(u,v),我们选其深度大的节点,把边权扔给它.因为这是树,所以每个点只有一个父亲,所以每个边权 ...
- 【LuoguP3038/[USACO11DEC]牧草种植Grass Planting】树链剖分+树状数组【树状数组的区间修改与区间查询】
模拟题,可以用树链剖分+线段树维护. 但是学了一个厉害的..树状数组的区间修改与区间查询.. 分割线里面的是转载的: ----------------------------------------- ...
- 树链剖分【p3038】[USACO11DEC]牧草种植Grass Planting
表示看不太清. 概括题意 树上维护区间修改与区间和查询. 很明显树剖裸题,切掉,细节处错误T了好久 TAT 代码 #include<cstdio> #include<cstdlib& ...
- 洛谷P3038 牧草种植Grass Planting
思路: 首先,这道题的翻译是有问题的(起码现在是),查询的时候应该是查询某一条路径的权值,而不是某条边(坑死我了). 与平常树链剖分题目不同的是,这道题目维护的是边权,而不是点权,那怎么办呢?好像有点 ...
- 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 b ...
随机推荐
- linux 下 docker-compose安装
docker和dockers-compose的版本兼容对照 以下是我的服务器的相关信息 linux版本 [root@izbp16fm097gaw3tdaog2wz bin]# cat /proc/ve ...
- 阿里云全国快递物流查询api接口
口地址: https://market.aliyun.com/products/56928004/cmapi021863.html?spm=5176.730005.productlist.d_cmap ...
- 图解Disruptor框架(一):初识Ringbuffer
图解Disruptor框架(一):初识Ringbuffer 概述 1. 什么是Disruptor?为什么是Disruptor? Disruptor是一个性能十分强悍的无锁高并发框架.在JUC并发包中, ...
- 前端之bootstrap
一.响应式介绍 众所周知,电脑.平板.手机的屏幕是差距很大的,假如在电脑上写好了一个页面,在电脑上看起来不错,但是如果放到手机上的话,那可能就会乱的一塌糊涂,这时候怎么解决呢?以前,可以再专门为手机定 ...
- How to setup multimedia on CentOS 7
You will need to also install the EPEL repository as nux-dextop depends on this for some of its pack ...
- Linux学习-系统基本设定
网络设定 (手动设定与 DHCP 自动取得) 网络其实是又可爱又麻烦的玩意儿,如果你是网络管理员,那么你必须要了解局域网络内的 IP, gateway, netmask 等参数,如果还想要连上 Int ...
- Linux学习-软件磁盘阵列 (Software RAID)
什么是 RAID 磁盘阵列全名是『 Redundant Arrays of Inexpensive Disks, RAID 』,英翻中的意思是:容错式廉价磁盘阵列.RAID 可以透过一个技术(软件或硬 ...
- HDU 4781 Assignment For Princess 构造
题意: 构造一个\(N(10 \leq N \leq 80)\)个顶点\(M(N+3 \leq M \leq \frac{N^2} {7})\)条边的有向图,要满足如下条件: 每条边有一个\([1,M ...
- Notepad++ 使用正则表达式查找替换字符串
最近在改一些别人写的PHP代码,由于之前的代码可维护性比较差,因此工作量比较多,因此想找一些高效的方式改一些有规律的代码. 比如: 我想将一些类似$rowss[MaxGetCash_num], ...
- 令人惊叹的sublime text 3 插件
1.ChineseLocalization------语言汉化.(新手必备) 2.SublimeTmpl------打开生成模板.(新手必备) 3.SublimeCodeIntel------代码自 ...