HDU 4358 Boring counting dfs序+莫队算法

题意：N个节点的有根树，每个节点有一个weight。有Q个查询，问在以u为根的子树中，有恰好出现了K次的weight有多少种。

 这是第一次写莫队算法，之前也只是偶有耳闻。

看了别人的代码打的，还是贴上来吧。

#pragma comment(linker, "/STACK:1000000000")
#include <iostream>
#include <cstdio>
#include <fstream>
#include <algorithm>
#include <cmath>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <cstring>
#include <map>
#include <stack>
#include <set>
#define LL long long
#define MAXN 100005
#define INF 0x3f3f3f3f
#define eps 1e-8
using namespace std;
int n, k, Q;
vector<int> p;
vector<int> G[MAXN];
int a[MAXN], sum[MAXN];
int L[MAXN], R[MAXN], res[MAXN];
int t, ans;
struct Node
{
    int left, right, id, pos;
    Node(int left = 0, int right = 0, int id = 0, int pos = 0):left(left), right(right), id(id), pos(pos){};
}m[MAXN];
bool compare(Node a, Node b)
{
    return (a.pos < b.pos ||(a.pos == b.pos && a.right < b.right));
}
void dfs(int x, int father)
{
    t++;
    L[x] = t;
    for(int i = 0; i < G[x].size(); i++){
        if(G[x][i] == father) continue;
        dfs(G[x][i], x);
    }
    R[x] = t;
}
void work(int x, int v)
{
    if(sum[x] == k){
        ans--;
    }
    sum[x] += v;
    if(sum[x] == k){
        ans++;
    }
}

int main()
{
    //freopen("in.txt", "r", stdin);
    int T;
    scanf("%d", &T);
    for(int cas = 1; cas <= T; cas++){
        scanf("%d%d", &n, &k);
        p.clear();
        for(int i = 1; i <= n; i++){
            scanf("%d", &a[i]);
            p.push_back(a[i]);
        }
        sort(p.begin(), p.end());
        for(int i = 1; i <= n; i++){
            a[i] = lower_bound(p.begin(), p.end(), a[i]) - p.begin();
        }
        int x, y;
        for(int i = 1; i <= n; i++){
            G[i].clear();
        }
        for(int i = 1; i < n; i++){
            scanf("%d%d", &x, &y);
            G[x].push_back(y);
            G[y].push_back(x);
        }
        t = 0;
        dfs(1, -1);
        scanf("%d", &Q);
        int u = sqrt(n);
        for(int i = 1; i <= Q; i++){
            scanf("%d", &x);
            m[i] = Node(L[x], R[x], i, L[x] / u);
        }
        sort(m + 1, m + Q + 1, compare);
        memset(sum, 0, sizeof(sum));
        int left = 1, right = 1;
        ans = 0;
        work(a[right], 1);
        for(int i = 1; i <= Q; i++){
            while(right < m[i].right){
                right++;
                work(a[right], 1);
            }
            while(right > m[i].right){
                work(a[right--], -1);
            }
            while(left < m[i].left){
                work(a[left++], -1);
            }
            while(left > m[i].left){
                left--;
                work(a[left], 1);
            }
            res[m[i].id] = ans;
        }
        printf("Case #%d:
", cas);
        for(int i = 1; i <= Q; i++){
            printf("%d
", res[i]);
        }
        if(cas != T){
            printf("
");
        }
    }
}