Tree and Queries CodeForces - 375D 树上莫队

Posted 2020-09-30 stupid_one

http://codeforces.com/problemset/problem/375/D

树莫队就是把树用dfs序变成线性的数组。 （原数组要根据dfs的顺序来变化）

然后和莫队一样的区间询问。

这题和普通莫队有点区别，他需要的不单单是统计区间元素种类个数，是区间元素种类个数 >= k[i]的个数。

考虑最简单的用bit维护，复杂度多了个log

观察到每次只需要 + 1  或者 -1

用一个数组sum[k]表示种类数大于等于k的ans

在numc[val]++之后，sum[numc[val]]++，表明这个种类出现次数是numc[val]次的贡献递增1

在num[val]--之前，就需要把sum[numc[val]]--，表明贡献减1了

#include <bits/stdc++.h>
#define ios ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;
const int maxn = 1e5 + 20;
struct Edge {
    int u, v, tonext;
}e[maxn * 2];
int first[maxn], num;
int c[maxn];
void addEdge(int u, int v) {
    ++num;
    e[num].u = u, e[num].v = v, e[num].tonext = first[u];
    first[u] = num;
}
int L[maxn], R[maxn], dfs_clock;
int color[maxn];
void dfs(int cur, int fa) {
    L[cur] = ++dfs_clock;
    color[dfs_clock] = c[cur];
    for (int i = first[cur]; i; i = e[i].tonext) {
        int v = e[i].v;
        if (fa == v) continue;
        dfs(v, cur);
    }
    R[cur] = dfs_clock;
}
int magic;
struct Node {
    int L, R, k, id;
    bool operator < (const struct Node & rhs) const {
        if (L / magic != rhs.L / magic) return L / magic < rhs.L / magic;
        else return R < rhs.R;
    }
}query[maxn];
int ans[maxn], numc[maxn];
int bit[maxn];
int lowbit(int x) {
    return x & (-x);
}
void add(int pos, int val) {
    while (pos) {
        bit[pos] += val;
        pos -= lowbit(pos);
    }
}
int ask(int pos) {
    int ans = 0;
    while (pos <= maxn - 20) {
        ans += bit[pos];
        pos += lowbit(pos);
    }
    return ans;
}
int suffix[maxn];
void work() {
    int n, m;
    cin >> n >> m;
    for (int i = 1; i <= n; ++i) cin >> c[i];
    for (int i = 1; i <= n - 1; ++i) {
        int u, v;
        scanf("%d%d", &u, &v);
        addEdge(u, v);
        addEdge(v, u);
    }
    dfs(1, 0);
    magic = (int)sqrt(n + 0.5);
    for (int i = 1; i <= m; ++i) {
        int which, k;
        scanf("%d%d", &which, &k);
        query[i].L = L[which], query[i].R = R[which], query[i].k = k;
        query[i].id = i;
    }
    sort(query + 1, query + 1 + m);
    int L = 1, R = 0;
    int temp = 0;
    for (int i = 1; i <= m; ++i) {
        while (R < query[i].R) {
            ++R;
            suffix[++numc[color[R]]]++;
        }
        while (R > query[i].R) {
            suffix[numc[color[R]]--]--;
            --R;
        }
        while (L < query[i].L) {
            suffix[numc[color[L]]--]--;
            L++;
        }
        while (L > query[i].L) {
            L--;
            suffix[++numc[color[L]]]++;
        }
        ans[query[i].id] = suffix[query[i].k];
    }
    for (int i = 1; i <= m; ++i) {
        printf("%d\n", ans[i]);
    }
}
int main() {
#ifdef local
    freopen("data.txt", "r", stdin);
//    freopen("data.txt", "w", stdout);
#endif
    work();
    return 0;
}