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