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Boring counting

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 98304/98304 K (Java/Others)
Total Submission(s): 1972    Accepted Submission(s): 561

Problem Description

In this problem we consider a rooted tree with N vertices. The vertices are numbered from 1 to N, and vertex 1 represents the root. There are integer weights on each vectice. Your task is to answer a list of queries, for each query, please tell us among all the vertices in the subtree rooted at vertice u, how many different kinds of weights appear exactly K times?

 

Input

The first line of the input contains an integer T( T<= 5 ), indicating the number of test cases.
For each test case, the first line contains two integers N and K, as described above. ( 1<= N <= 10⁵, 1 <= K <= N )
Then come N integers in the second line, they are the weights of vertice 1 to N. ( 0 <= weight <= 10⁹ )
For next N-1 lines, each line contains two vertices u and v, which is connected in the tree.
Next line is a integer Q, representing the number of queries. (1 <= Q <= 10⁵)
For next Q lines, each with an integer u, as the root of the subtree described above.

 

Output

For each test case, output "Case #X:" first, X is the test number. Then output Q lines, each with a number -- the answer to each query.

Seperate each test case with an empty line.

 

Sample Input

13 11 2 21 21 33213

 

Sample Output

Case #1:111

树上的查询。

思路：首先dfs对节点进行编号，转化成线性的，然后离线处理。更新的时候注意，下面举个例子：

0  1  2  3  4

2  2  2  2  2

当处理到位置4的时候，要对2位置-2,1位置+1,3位置+1，因为2位置先前是1，-2的目的是，当查询（2,4）是结果是零，同理，原来1位置是-1，+1后变成0，这样同样保证查询的正确性。

#include<iostream>#include<cstdio>#include<cstring>#include<vector>#include<cmath>#include<queue>#include<stack>#include<map>#include<set>#include<algorithm>using namespace std;const int maxn=100100;int N,K,Q,idx;int data1[maxn],data2[maxn];int low[maxn],high[maxn],ans[maxn];map<int,int> m;vector<int> g[maxn];vector<int> pos[maxn];struct QU{    int s,e,id;    bool operator < (const QU & a)const    {        return e<a.e;    }}qu[maxn];struct IntervalTree{    int sum[maxn<<3];    void build(){memset(sum,0,sizeof(sum));}    void update(int o,int l,int r,int pos,int val)    {        sum[o]+=val;        if(l==r)return ;        int mid=(l+r)>>1;        if(pos<=mid)update(o<<1,l,mid,pos,val);        else update(o<<1|1,mid+1,r,pos,val);    }    int query(int o,int l,int r,int q1,int q2)    {        if(q1<=l&&r<=q2)return sum[o];        int mid=(l+r)>>1;        int ans=0;        if(q1<=mid)ans+=query(o<<1,l,mid,q1,q2);        if(q2>mid)ans+=query(o<<1|1,mid+1,r,q1,q2);        return ans;    }}tree;void dfs(int u,int fa){    low[u]=++idx;    data2[idx]=data1[u];    int len=g[u].size();    for(int i=0;i<len;i++)    {        int v=g[u][i];        if(fa==v)continue;        dfs(v,u);    }    high[u]=idx;}void init(){    m.clear();    for(int i=0;i<=N;i++)    {        g[i].clear();        pos[i].clear();    }    idx=0;}int main(){    int T,cas=1,first=1;    scanf("%d",&T);    while(T--)    {        if(first)first=0;        else printf("\n");        scanf("%d%d",&N,&K);        int dataid=0;        init();        for(int i=1;i<=N;i++)        {            scanf("%d",&data1[i]);            if(m.find(data1[i])==m.end())                m[data1[i]]=dataid++;        }        for(int i=1;i<N;i++)        {            int u,v;            scanf("%d%d",&u,&v);            g[u].push_back(v);        }        dfs(1,-1);        scanf("%d",&Q);        for(int i=1;i<=Q;i++)        {            int s;            scanf("%d",&s);            qu[i].s=low[s];            qu[i].e=high[s];            qu[i].id=i;        }        sort(qu+1,qu+1+Q);        tree.build();        int num=1;        for(int i=1;i<=N;i++)        {            int tmp=m[data2[i]];            pos[tmp].push_back(i);            int cnt=pos[tmp].size();            if(cnt>=K)            {                if(cnt>K)tree.update(1,1,idx,pos[tmp][cnt-K-1],-2);                if(cnt>K+1)tree.update(1,1,idx,pos[tmp][cnt-K-2],1);                tree.update(1,1,idx,pos[tmp][cnt-K],1);            }            while(num<=Q&&qu[num].e==i)            {                ans[qu[num].id]=tree.query(1,1,idx,qu[num].s,qu[num].e);                num++;            }        }        printf("Case #%d:\n",cas++);        for(int i=1;i<Q;i++)printf("%d\n",ans[i]);        printf("%d\n",ans[Q]);    }    return 0;}