hdu 4358 Boring counting 离散化+dfs序+莫队算法

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4358

题意:以1为根节点含有N(N <= 1e5)个结点的树，每个节点有一个权值(weight <= 1e9)。之后有m(m <= 1e5)次查询，每次查询以节点u为子树的树中，权值出现k次的权值有多少个？

Sample Input

1

3 1 (n,k)

1 2 2

1 2

1 3

3 (m)

2 1 3

 

Sample Output

Case #1:

1

1

1

 

思路:建好树之后，dfs得到L[x],R[x].即将树形结构变成了线性的结构。并且由于问的是权值出现k次的权值有多少个，和权值的大小无关。就可以离散化，在dfs中把节点重新排序，并且重新赋值.(我们只关心dfs序之后的id和val);这样之后直接跑莫队即可；

细节:至于离散化，直接使用了vector和lower_bound()来合并相同的值；之后输入时建好查询的q[]即可.

#pragma comment(linker, "/STACK:1024000000")
#include<bits/stdc++.h>
using namespace std;
#define rep0(i,l,r) for(int i = (l);i < (r);i++)
#define rep1(i,l,r) for(int i = (l);i <= (r);i++)
#define rep_0(i,r,l) for(int i = (r);i > (l);i--)
#define rep_1(i,r,l) for(int i = (r);i >= (l);i--)
#define MS0(a) memset(a,0,sizeof(a))
#define MS1(a) memset(a,-1,sizeof(a))
#define MSi(a) memset(a,0x3f,sizeof(a))
#define inf 0x3f3f3f3f
#define lson l, m, rt << 1
#define rson m+1, r, rt << 1|1
typedef pair<int,int> PII;
#define A first
#define B second
#define MK make_pair
#define pb push_back
typedef __int64 ll;
template<typename T>
void read1(T &m)
{
    T x=,f=;char ch=getchar();
    while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
    while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
    m = x*f;
}
template<typename T>
void read2(T &a,T &b){read1(a);read1(b);}
template<typename T>
void read3(T &a,T &b,T &c){read1(a);read1(b);read1(c);}
template<typename T>
void out(T a)
{
    if(a>) out(a/);
    putchar(a%+'');
}
const int M = ;
int head[M],tot,a[M],dfs_clock;
int T,kase = ,i,j,k,n,m,top,cnt[M];
struct Edge{
    int to,w,Next;
    Edge(){}
    Edge(int to,int w,int Next):to(to),w(w),Next(Next){}
}e[M<<];
inline void ins(int u,int v,int w = )
{
    e[++tot] = Edge{v,w,head[u]};
    head[u] = tot;
}
vector<int> p;
void init()
{
    dfs_clock = tot = ;
    MS0(head);p.clear();
    MS0(cnt);
}
int L[M],R[M],val[M],ans[M];
void dfs(int u,int pre)
{
    L[u] = ++dfs_clock;
    val[dfs_clock] = a[u];//关系的只是dfs序之后的序号;
    for(int d = head[u];d;d = e[d].Next){
        int v = e[d].to;
        if(v == pre) continue;
        dfs(v,u);
    }
    R[u] = dfs_clock;
}
struct data{
    int l,r,id,block;
    data(){}
    data(int l,int r,int id,int block):l(l),r(r),id(id),block(block){}
}q[M];
bool cmp(const data& a,const data& b)
{
    return a.block - b.block?a.block < b.block:a.r < b.r;
}
void update(int pos,int add)
{
    if(cnt[val[pos]] == k) top--;
    else if(cnt[val[pos]] + add == k) top++;
    cnt[val[pos]] += add;
}
void solve()
{
    top = ;
    for(int i = ,l = ,r = ;i <= m;i++){
        while(r < q[i].r) update(++r,);
        while(r > q[i].r) update(r--,-);
        while(l < q[i].l) update(l++,-);
        while(l > q[i].l) update(--l,);
        ans[q[i].id] = top;
    }
}
int main()
{
    read1(T);
    while(T--){
        init();
        read2(n,k);
        rep1(i,,n) read1(a[i]),p.pb(a[i]);
        sort(p.begin(),p.end());
        rep1(i,,n) //离散化;
            a[i] = lower_bound(p.begin(),p.end(),a[i]) - p.begin();
        int u,v;
        rep0(i,,n){
            read2(u,v);
            ins(u,v);ins(v,u);
        }
        dfs(,-);
        int block = sqrt(n);
        read1(m);
        rep1(i,,m){
            read1(u);
            q[i] = data(L[u],R[u],i,L[u]/block);
        }
        sort(q+,q++m,cmp);
        solve();
        if(kase) puts("");
        printf("Case #%d:\n",++kase);
        rep1(i,,m){
            out(ans[i]);
            puts("");
        }
    }
    return ;
}