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。

1 #pragma comment(linker, "/STACK:1000000000")
2 #include <iostream>
3 #include <cstdio>
4 #include <fstream>
5 #include <algorithm>
6 #include <cmath>
7 #include <deque>
8 #include <vector>
9 #include <queue>
10 #include <string>
11 #include <cstring>
12 #include <map>
13 #include <stack>
14 #include <set>
15 #define LL long long
16 #define MAXN 100005
17 #define INF 0x3f3f3f3f
18 #define eps 1e-8
19 using namespace std;
20 int n, k, Q;
21 vector<int> p;
22 vector<int> G[MAXN];
23 int a[MAXN], sum[MAXN];
24 int L[MAXN], R[MAXN], res[MAXN];
25 int t, ans;
26 struct Node
27 {
28     int left, right, id, pos;
29     Node(int left = 0, int right = 0, int id = 0, int pos = 0):left(left), right(right), id(id), pos(pos){};
30 }m[MAXN];
31 bool compare(Node a, Node b)
32 {
33     return (a.pos < b.pos ||(a.pos == b.pos && a.right < b.right));
34 }
35 void dfs(int x, int father)
36 {
37     t++;
38     L[x] = t;
39     for(int i = 0; i < G[x].size(); i++){
40         if(G[x][i] == father) continue;
41         dfs(G[x][i], x);
42     }
43     R[x] = t;
44 }
45 void work(int x, int v)
46 {
47     if(sum[x] == k){
48         ans--;
49     }
50     sum[x] += v;
51     if(sum[x] == k){
52         ans++;
53     }
54 }
55
56 int main()
57 {
58     //freopen("in.txt", "r", stdin);
59     int T;
60     scanf("%d", &T);
61     for(int cas = 1; cas <= T; cas++){
62         scanf("%d%d", &n, &k);
63         p.clear();
64         for(int i = 1; i <= n; i++){
65             scanf("%d", &a[i]);
66             p.push_back(a[i]);
67         }
68         sort(p.begin(), p.end());
69         for(int i = 1; i <= n; i++){
70             a[i] = lower_bound(p.begin(), p.end(), a[i]) - p.begin();
71         }
72         int x, y;
73         for(int i = 1; i <= n; i++){
74             G[i].clear();
75         }
76         for(int i = 1; i < n; i++){
77             scanf("%d%d", &x, &y);
78             G[x].push_back(y);
79             G[y].push_back(x);
80         }
81         t = 0;
82         dfs(1, -1);
83         scanf("%d", &Q);
84         int u = sqrt(n);
85         for(int i = 1; i <= Q; i++){
86             scanf("%d", &x);
87             m[i] = Node(L[x], R[x], i, L[x] / u);
88         }
89         sort(m + 1, m + Q + 1, compare);
90         memset(sum, 0, sizeof(sum));
91         int left = 1, right = 1;
92         ans = 0;
93         work(a[right], 1);
94         for(int i = 1; i <= Q; i++){
95             while(right < m[i].right){
96                 right++;
97                 work(a[right], 1);
98             }
99             while(right > m[i].right){
100                 work(a[right--], -1);
101             }
102             while(left < m[i].left){
103                 work(a[left++], -1);
104             }
105             while(left > m[i].left){
106                 left--;
107                 work(a[left], 1);
108             }
109             res[m[i].id] = ans;
110         }
111         printf("Case #%d:\n", cas);
112         for(int i = 1; i <= Q; i++){
113             printf("%d\n", res[i]);
114         }
115         if(cas != T){
116             printf("\n");
117         }
118     }
119 }