HDU 4358 Boring counting（莫队+DFS序+离散化）

Boring counting

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 98304/98304 K (Java/Others)
Total Submission(s): 2811    Accepted Submission(s): 827

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

1

3 1

1 2 2

1 2

1 3

3

2

1

3

 

Sample Output

Case #1:

1

1

1

题目链接：HDU 4358

把DFS序和莫队算法结合了起来，前两发杯具PE，如果用过DFS序配合线段树的话就大概能知道怎么做了，对于一颗子树的询问显然DFS序是很适合的，然后这样就得到了每一个点所管理的区间[L,R]，那莫队移动的时候怎么判断是否遇到了某一个原树上的节点呢？显然用先序遍历的方式来得到某一个点管理的区间，那么L这个点必定是子树树根的位置，若这个点管理的位置是[L,R]那么实际上这个子树根点的值就是arr[L]，不过由于每一个区间都是满点的，就映射成val[timeorder]=arr[u]，其中u是某次dfs时的起点。

除此之外还要判断当前减掉的数字是从k减到k-1还是k+1减到k，加上的同理，最后每一个case之间换一行，结尾不换行Orz……另外最近在玩C++11的匿名函数，在sort里随便用下玩

代码：

#include <stdio.h>
#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define CLR(arr,val) memset(arr,val,sizeof(arr))
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
typedef pair<int,int> pii;
typedef long long LL;
const double PI=acos(-1.0);
const int N=1e5+7;
struct edge
{
    int to;
    int pre;
};
struct info
{
    int l,r;
    int id,b;
};
info Q[N];
edge E[N];
int head[N],tot;
int L[N],R[N],order,val[N];
int arr[N],cnt[N],ans[N];
vector<int>vec;

void init()
{
    CLR(head,-1);
    tot=0;
    CLR(L,0);
    CLR(R,0);
    order=0;
    CLR(val,0);
    CLR(cnt,0);
    vec.clear();
}
inline void add(int s,int t)
{
    E[tot].to=t;
    E[tot].pre=head[s];
    head[s]=tot++;
}
void dfs(const int &now,const int &pre)
{
    L[now]=++order;
    val[order]=arr[now];
    for (int i=head[now]; ~i; i=E[i].pre)
    {
        int v=E[i].to;
        //if(v!=pre)
        dfs(v,now);
    }
    R[now]=order;
}
int main(void)
{
    int tcase,n,k,i,a,b,m,rt;
    scanf("%d",&tcase);
    for (int q=1; q<=tcase; ++q)
    {
        init();
        scanf("%d%d",&n,&k);
        for (i=1; i<=n; ++i)
        {
            scanf("%d",&arr[i]);
            vec.push_back(arr[i]);
        }
        sort(vec.begin(),vec.end());
        vec.erase(unique(vec.begin(),vec.end()),vec.end());
        for (i=1; i<=n; ++i)
            arr[i]=lower_bound(vec.begin(),vec.end(),arr[i])-vec.begin();
        for (i=0; i<n-1; ++i)
        {
            scanf("%d%d",&a,&b);
            add(a,b);
        }
        dfs(1,-1);
        scanf("%d",&m);
        int unit=(int)sqrt(1.0*n);
        for (i=0; i<m; ++i)
        {
            scanf("%d",&rt);
            Q[i].l=L[rt];
            Q[i].r=R[rt];
            Q[i].id=i;
            Q[i].b=Q[i].l/unit;
        }
        sort(Q,Q+m,[&](const info &x,const info &y){return (x.b==y.b&&x.r<y.r)||x.b<y.b;});
        int l=1,r=0,temp=0;
        for (i=0; i<m; ++i)
        {
            while (l<Q[i].l)
            {
                --cnt[val[l]];
                if(cnt[val[l]]==k)
                    ++temp;
                else if(cnt[val[l]]==k-1)
                    --temp;
                ++l;
            }
            while (l>Q[i].l)
            {
                --l;
                ++cnt[val[l]];
                if(cnt[val[l]]==k)
                    ++temp;
                else if(cnt[val[l]]==k+1)
                    --temp;
            }
            while (r<Q[i].r)
            {
                ++r;
                ++cnt[val[r]];
                if(cnt[val[r]]==k)
                    ++temp;
                else if(cnt[val[r]]==k+1)
                    --temp;
            }
            while (r>Q[i].r)
            {
                --cnt[val[r]];
                if(cnt[val[r]]==k-1)
                    --temp;
                else if(cnt[val[r]]==k)
                    ++temp;
                --r;
            }
            ans[Q[i].id]=temp;
        }
        printf("Case #%d:\n",q);
        for (i=0; i<m; ++i)
            printf("%d\n",ans[i]);
        if(q!=tcase)
            putchar('\n');
    }
    return 0;
}