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

输入样例#1:

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
输出样例#1:

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

  1. 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 ...

  2. 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 ...

  3. 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 ...

  4. 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 ...

  5. [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 ...

  6. 【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: ----------------------------------------- ...

  7. 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& ...

  8. 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 ...

  9. 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

  1. PHP使用FTP上传文件到服务器(实战篇)

    我们在做开发的过程中,上传文件肯定是避免不了的,平常我们的程序和上传的文件都在一个服务器上,我们也可以使用第三方sdk上传文件,但是文件在第三方服务器上.现在我们使用PHP的ftp功能把文件上传到我们 ...

  2. 再生龙备份还原linux系统

    相关下载: Clonezilla再生龙:http://sourceforge.net/projects/clonezilla/files/clonezilla_live_stable/ tuxboot ...

  3. poj3617 best cow line(贪心题)

    Best Cow Line Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 32687   Accepted: 8660 De ...

  4. 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 ...

  5. CodeForces 379F 树的直径 New Year Tree

    题意:每次操作新加两个叶子节点,每次操作完以后询问树的直径. 维护树的直径的两个端点U,V,每次计算一下新加进来的叶子节点到U,V两点的距离,如果有更长的就更新. 因为根据树的直径的求法,若出现新的直 ...

  6. python基础学习笔记——反射

    对编程语言比较熟悉的朋友,应该知道“反射”这个机制.Python作为一门动态语言,当然不会缺少这一重要功能.然而,在网络上却很少见到有详细或者深刻的剖析论文.下面结合一个web路由的实例来阐述pyth ...

  7. loj2026 「JLOI / SHOI2016」成绩比较

    orz #include <iostream> #include <cstdio> using namespace std; typedef long long ll; int ...

  8. xml报错“cvc-complex-type.2.4.c: The matching wildcard is strict, but no declaration can be found for element”

    配置使用dubbo时,xml报错“cvc-complex-type.2.4.c: The matching wildcard is strict, but no declaration can be ...

  9. hdu3667

    Transportation Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)To ...

  10. [luoguP2463] [SDOI2008]Sandy的卡片(后缀数组 + st表)

    传送门 很容易想到,题目中的相同是指差分数组相同. 那么可以把差分数组连起来,中间加上一个没有出现过的且字典序小的数 双指针移动,用st表维护height数组中的最小值. 当然用单调队列应该也可以且更 ...