COT2 - Count on a tree II
http://www.spoj.com/problems/COT2/
You are given a tree with N nodes. The tree nodes are numbered from 1 to N. Each node has an integer weight.
We will ask you to perform the following operation:
- u v : ask for how many different integers that represent the weight of nodes there are on the path from u tov.
Input
In the first line there are two integers N and M. (N <= 40000, M <= 100000)
In the second line there are N integers. The i-th integer denotes the weight of the i-th node.
In the next N-1 lines, each line contains two integers u v, which describes an edge (u, v).
In the next M lines, each line contains two integers u v, which means an operation asking for how many different integers that represent the weight of nodes there are on the path from u to v.
Output
For each operation, print its result.
Example
Input: 8 2 105 2 9 3 8 5 7 7 1 2 1 3 1 4 3 5 3 6 3 7 4 8 2 5 7 8
Output: 4 4
题意：问树上两点间有多少不同的权值
树上莫队
开始狂T，发现自己竟是按节点编号划分的块！！
dfs分块。。
#include<cmath>
#include<cstdio>
#include<algorithm>
using namespace std;
#define N 40001
#define M 100001
int n,m,siz,tmp;
int hash[N],key[N];
int front[N],to[N*2],nxt[N*2],tot;
int fa[N],deep[N],id[N],son[N],bl[N],block[N];
bool vis[N];
int sum[N],ans[M];
struct node
{
int l,r,id;
bool operator < (node p) const
{
if(block[l]!=block[p.l]) return block[l]<block[p.l];
return block[r]<block[p.r];
}
}e[M];
int read(int &x)
{
x=0; char c=getchar();
while(c<'0'||c>'9') c=getchar();
while(c>='0'&&c<='9') { x=x*10+c-'0'; c=getchar(); }
}
void add(int x,int y)
{
to[++tot]=y; nxt[tot]=front[x]; front[x]=tot;
to[++tot]=x; nxt[tot]=front[y]; front[y]=tot;
}
void dfs(int x)
{
son[x]++;
for(int i=front[x];i;i=nxt[i])
{
if(to[i]==fa[x]) continue;
deep[to[i]]=deep[x]+1;
fa[to[i]]=x;
dfs(to[i]);
son[x]+=son[to[i]];
}
}
void dfs2(int x,int top)
{
id[x]=++tot;
bl[x]=top;
block[x]=(tot-1)/siz+1;
int y=0;
for(int i=front[x];i;i=nxt[i])
{
if(to[i]==fa[x]) continue;
if(son[to[i]]>son[y]) y=to[i];
}
if(!y) return;
dfs2(y,top);
for(int i=front[x];i;i=nxt[i])
{
if(to[i]==fa[x]||to[i]==y) continue;
dfs2(to[i],to[i]);
}
}
void point(int u)
{
if(vis[u]) tmp-=(!--sum[hash[u]]);
else tmp+=(++sum[hash[u]]==1);
vis[u]^=1;
}
void path(int u,int v)
{
while(u!=v)
{
if(deep[u]>deep[v]) point(u),u=fa[u];
else point(v),v=fa[v];
}
}
int get_lca(int u,int v)
{
while(bl[u]!=bl[v])
{
if(deep[bl[u]]<deep[bl[v]]) swap(u,v);
u=fa[bl[u]];
}
return deep[u]<deep[v] ? u : v;
}
int main()
{
read(n);read(m); siz=sqrt(n);
for(int i=1;i<=n;i++) read(key[i]),hash[i]=key[i];
sort(key+1,key+n+1);
key[0]=unique(key+1,key+n+1)-(key+1);
for(int i=1;i<=n;i++) hash[i]=lower_bound(key+1,key+key[0]+1,hash[i])-key;
int x,y;
for(int i=1;i<n;i++)
{
read(x); read(y);
add(x,y);
}
tot=0;
dfs(1);
dfs2(1,1);
for(int i=1;i<=m;i++)
{
read(e[i].l); read(e[i].r);
e[i].id=i;
}
sort(e+1,e+m+1);
int L=1,R=1,lca;
for(int i=1;i<=m;i++)
{
if(id[e[i].l]>id[e[i].r]) swap(e[i].l,e[i].r);
path(L,e[i].l);
path(R,e[i].r);
lca=get_lca(e[i].l,e[i].r);
point(lca);
ans[e[i].id]=tmp;
point(lca);
L=e[i].l; R=e[i].r;
}
for(int i=1;i<=m;i++) printf("%d\n",ans[i]);
}
作者：xxy
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