http://www.elijahqi.win/2018/03/01/codeforces-375d/
题意翻译
给出一棵 n 个结点的树，每个结点有一个颜色 c i 。 询问 q 次，每次询问以 v 结点为根的子树中，出现次数 ≥k 的颜色有多少种。
感谢@elijahqi 提供的翻译
题目描述
You have a rooted tree consisting of
n
n vertices. Each vertex of the tree has some color. We will assume that the tree vertices are numbered by integers from 1 to
n
n . Then we represent the color of vertex
v
v as
c_{v}
cv . The tree root is a vertex with number 1.
In this problem you need to answer to
m
m queries. Each query is described by two integers
v_{j},k_{j}
vj,kj . The answer to query
v_{j},k_{j}
vj,kj is the number of such colors of vertices
x
x , that the subtree of vertex
v_{j}
vj contains at least
k_{j}
kj vertices of color
x
x .
You can find the definition of a rooted tree by the following link: http://en.wikipedia.org/wiki/Tree_(graph_theory).
输入输出格式
输入格式：
The first line contains two integers
n
n and
m
m (2<=n<=105;1<=m<=105)
(
2
<=
n
<=
10
5
;
1
<=
m
<=
10
5
)
c_{1},c_{2},…,c_{n}
c1,c2,…,cn
(1<=c_{i}<=10^{5})
(1<=ci<=105) . The next
n-1
n−1 lines contain the edges of the tree. The
i
i -th line contains the numbers
a_{i},b_{i}
ai,bi (1<=ai,bi<=n;ai≠bi)
(
1
<=
a
i
,
b
i
<=
n
;
a
i
≠
b
i
)
Next
m
m lines contain the queries. The
j
j -th line contains two integers
v_{j},k_{j}
vj,kj (1<=vj<=n;1<=kj<=105)
(
1
<=
v
j
<=
n
;
1
<=
k
j
<=
10
5
)
输出格式：
Print
m
m integers — the answers to the queries in the order the queries appear in the input.
输入输出样例
输入样例#1： 复制
8 5
1 2 2 3 3 2 3 3
1 2
1 5
2 3
2 4
5 6
5 7
5 8
1 2
1 3
1 4
2 3
5 3
输出样例#1： 复制
2
2
1
0
1
输入样例#2： 复制
4 1
1 2 3 4
1 2
2 3
3 4
1 1
输出样例#2： 复制
4
说明
A subtree of vertex
v
v in a rooted tree with root
r
r is a set of vertices u:dist(r,v)+dist(v,u)=dist(r,u)
u
:
d
i
s
t
(
r
,
v
)
+
d
i
s
t
(
v
,
u
)
=
d
i
s
t
(
r
,
u
)
dist(x,y)
dist(x,y) is the length (in edges) of the shortest path between vertices
x
x and
y
y .
考虑子树一定是dfs序列上一个连续的区间 那么每次对子树操作总相当于针对dfs序上的一段连续区间操作
就可以莫队了 设ans表示出现次数超过i次的颜色数 f[i]表示i号颜色出现的次数 Ans相对应的记录每个询问的答案 那么考虑每个颜色出现一定是从0~xx 那么我每出现一次 我即对ans进行修改即可 并不需要区间修改 然后相应的减去的时候 也是这时候的出现次数-1 然后对应的ans数组再减1即可
#include<cmath>
#include<cstdio>
#include<algorithm>
#define N 110000
using namespace std;
inline char gc(){
static char now[1<<16],*S,*T;
if (T==S){T=(S=now)+fread(now,1,1<<16,stdin);if (T==S) return EOF;}
return *S++;
}
inline int read(){
int x=0,f=1;char ch=gc();
while(ch<'0'||ch>'9') {if (ch=='-') f=-1;ch=gc();}
while(ch<='9'&&ch>='0') x=x*10+ch-'0',ch=gc();
return x*f;
}
struct node1{
int y,next;
}data[N<<1];
struct node{
int l,r,id,bl,k;
}qr[N];bool visit[N];
int num,h[N],in[N],out[N],w[N],c[N],ans[N],Ans[N],f[N],nn,n,m;
inline bool cmp(const node &a,const node &b){
return a.bl==b.bl?a.r<b.r:a.bl<b.bl;
}
inline void dfs(int x,int fa){
in[x]=++num;w[num]=c[x];
for (int i=h[x];i;i=data[i].next){
int y=data[i].y;if (y==fa) continue;dfs(y,x);
}out[x]=num;
}
inline void update(int x){
if (visit[x]) --ans[f[w[x]]--];
else ++ans[++f[w[x]]];
visit[x]^=1;
}
int main(){
// freopen("cf375d.in","r",stdin);
n=read();m=read();for (int i=1;i<=n;++i) c[i]=read();
for (int i=1;i<n;++i){
int x=read(),y=read();
data[++num].y=y;data[num].next=h[x];h[x]=num;
data[++num].y=x;data[num].next=h[y];h[y]=num;
}num=0;dfs(1,1);nn=sqrt(n);
for (int i=1;i<=m;++i) {
qr[i].id=i;int x=read();qr[i].l=in[x];qr[i].r=out[x];qr[i].k=read();
qr[i].bl=(qr[i].l-1)/nn+1;
}sort(qr+1,qr+m+1,cmp);
int cl=1,cr=0;
for (int i=1;i<=m;++i){
int l=qr[i].l,r=qr[i].r;
while(cl<l) update(cl++);
while(cl>l) update(--cl);
while(cr>r) update(cr--);
while(cr<r) update(++cr);
Ans[qr[i].id]=ans[qr[i].k];
}
for (int i=1;i<=m;++i) printf("%d\n",Ans[i]);
return 0;
}
