BZOJ3351: [ioi2009]Regions(根号分治)

题意

sol

很神仙的题

我们考虑询问(a, b)(a是b的祖先)，直接对b根号分治

如果b的出现次数\(< \sqrt{n}\)，我们可以直接对每个b记录下与它有关的询问，这样每个询问至多扫\(\sqrt{n}\)个点即可知道答案，那么dfs的时候暴力统计答案即可，复杂度\(q\sqrt{n}\)

如果b的出现次数\(> \sqrt{n}\)，显然这样的b最多只有\(\sqrt{n}\)个，也就是说在询问中最多会有\(\sqrt{n}\)个这样的b，那么我们可以对每个a，暴力统计，复杂度\(n\sqrt{n}\)

然后用天天爱跑步那题的差分技巧搞一下就行了

代码十分好写~

#include<bits/stdc++.h> 
#define pair pair<int, int>
#define mp(x, y) make_pair(x, y)
#define fi first
#define se second
//#define int long long 
#define ll long long 
#define fin(x) {freopen(#x".in","r",stdin);}
#define fout(x) {freopen(#x".out","w",stdout);}
using namespace std;
const int maxn = 1e6 + 10;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int n, r, q, base;
vector<int> v[maxn];
vector<pair> a1[maxn], a2[maxn];
int r[maxn], fa[maxn], ti[maxn], ha[maxn], ans[maxn];
void dfs1(int x) {
    for(auto &a : a1[r[x]]) ans[a.se] += ha[a.fi]; ha[r[x]]++;
    for(auto &to : v[x]) dfs1(to); ha[r[x]]--;
}
void dfs2(int x) {
    for(auto &b: a2[r[x]]) ans[b.se] -= ha[b.fi];
    for(auto &to: v[x]) dfs2(to);
    for(auto &b: a2[r[x]]) ans[b.se] += ha[b.fi];
    ha[r[x]]++;
}
signed main() {
//  fin(9); fout(b);
    n = read(); r = read(); q = read(); base = sqrt(n);
    r[1] = read(); ti[r[1]]++;
    for(int i = 2; i <= n; i++) {
        int f = read(); r[i] = read();
        v[f].push_back(i);
        ti[r[i]]++;
    }
    for(int i = 1; i <= q; i++) {
        int a = read(), b = read();
        if(ti[b] < base) {
            a1[b].push_back({a, i});
        } else {
            a2[a].push_back({b, i});
        }
    }
    dfs1(1);
    memset(ha, 0, sizeof(ha));
    dfs2(1);
    for(int i = 1; i <= q; i++) printf("%d\n", ans[i]);
    return 0;
}