题解 CF375D 【Tree and Queries】

首先，子树上的查询问题可以通过$DFS$序转为序列问题

再一看，没有修改，可以离线，这不就是莫队吗？

我们用$sum_i$表示出现次数$\geq i$的个数

用$val_i$表示第$i$种颜色的出现次数

那么每次修改时只要$O(1)$修改$sum$和$val$即可

详见代码

 #include <bits/stdc++.h>
 const int MaxN = ;
 struct node
 {
     int val, dfn, r, id;
 };
 struct query
 {
     int l, r;
     int pos, id, k;
 };
 struct edge
 {
     int next, to;
 };
 node a[MaxN];
 query q[MaxN];
 edge e[MaxN << ];
 int n, m, cnt, dfscnt, size;
 int head[MaxN], ans[MaxN], sum[MaxN], val[MaxN];
 inline int comp(node a, node b) { return a.dfn < b.dfn; }
 inline int cmp(query a, query b)
 {
     if (a.pos != b.pos)
         return a.pos < b.pos;
     return a.r < b.r;
 }
 inline void add_edge(int u, int v)
 {
     ++cnt;
     e[cnt].to = v;
     e[cnt].next = head[u];
     head[u] = cnt;
 }
 inline void dfs(int u)
 {
     a[u].dfn = ++dfscnt;
     for (int i = head[u]; i; i = e[i].next)
     {
         int v = e[i].to;
         if (!a[v].dfn)
             dfs(v);
     }
     a[u].r = dfscnt;
 }
 inline int read()
 {
     int x = ;
     char ch = getchar();
     while (ch > '' || ch < '')
         ch = getchar();
     while (ch <= '' && ch >= '')
         x = (x << ) + (x << ) + (ch ^ ), ch = getchar();
     return x;
 }
 inline void add(int x) { ++val[a[x].val], ++sum[val[a[x].val]]; }
 inline void del(int x) { --sum[val[a[x].val]], --val[a[x].val]; }
 inline void solve()
 {
     int l = , r = ;
     for (int i = ; i <= m; i++)
     {
         while (l > q[i].l)
             --l, add(l);
         while (r < q[i].r)
             ++r, add(r);
         while (l < q[i].l)
             del(l), ++l;
         while (r > q[i].r)
             del(r), --r;
         ans[q[i].id] = sum[q[i].k];
     }
 }
 int main()
 {
     n = read(), m = read();
     size = pow(n, 0.55);
     for (int i = ; i <= n; i++)
         a[i].val = read(), a[i].id = i;
     for (int i = ; i <= n - ; i++)
     {
         int u = read(), v = read();
         add_edge(u, v);
         add_edge(v, u);
     }
     dfs();
     for (int i = ; i <= m; i++)
     {
         int v, k;
         v = read(), k = read();
         q[i].l = a[v].dfn, q[i].r = a[v].r, q[i].k = k;
         q[i].id = i, q[i].pos = (q[i].l - ) / size + ;
     }
     std::sort(q, q + m + , cmp);
     std::sort(a + , a + n + , comp);
     solve();
     for (int i = ; i <= m; i++)
         printf("%d\n", ans[i]);
     return ;
 }