Codeforces 1039D You Are Given a Tree (看题解)

You Are Given a Tree

感觉是个套路。。 怎么好像我没怎么见过啊。

k * t  <= n 类似于这种， 对所有 k 求满足条件 t 的最大值， 那么答案不同的数量只有根号个。

如果随 k 的变化单调的话就可以二分优化啦。

#include<bits/stdc++.h>
#define LL long long
#define LD long double
#define ull unsigned long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ALL(x) (x).begin(), (x).end()
#define fio ios::sync_with_stdio(false); cin.tie(0);

using namespace std;

const int N = 1e5 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double PI = acos(-1);

template<class T, class S> inline void add(T& a, S b) {a += b; if(a >= mod) a -= mod;}
template<class T, class S> inline void sub(T& a, S b) {a -= b; if(a < 0) a += mod;}
template<class T, class S> inline bool chkmax(T& a, S b) {return a < b ? a = b, true : false;}
template<class T, class S> inline bool chkmin(T& a, S b) {return a > b ? a = b, true : false;}

const int B = 1000;

int n, k, cur, ans[N], fa[N];
vector<int> G[N];
int id[N], nid;
int dp[N];
int mx[N][2];

int calc(int k) {
    if(~ans[k]) return ans[k];
    for(int o = 1; o <= n; o++) {
        int u = id[o];
        dp[u] = mx[u][0] = mx[u][1] = 0;
        for (auto& v : G[u]) {
            if(v == fa[u]) continue;
            dp[u] += dp[v];
            if(mx[v][0] >= mx[u][0]) mx[u][1] = mx[u][0], mx[u][0] = mx[v][0];
            else if(mx[v][0] > mx[u][1]) mx[u][1] = mx[v][0];
        }
        if(mx[u][0] + mx[u][1] + 1 >= k) {
            dp[u]++;
            mx[u][0] = mx[u][1] = 0;
        } else {
            mx[u][0]++;
            mx[u][1]++;
        }
    }
    return ans[k] = dp[1];
}

void dfs(int u, int f) {
    fa[u] = f;
    for(auto& v : G[u]) {
        if(v == f) continue;
        dfs(v, u);
    }
    id[++nid] = u;
}

void solve(int l, int r) {
    if(l > r) return;
    if(calc(l) == calc(r)) {
        for(int i = l + 1; i < r; i++) ans[i] = ans[l];
        return;
    }
    int mid = l + r >> 1;
    solve(l, mid); solve(mid + 1, r);
}

int main() {
    memset(ans, -1, sizeof(ans));
    scanf("%d", &n);
    for(int i = 1; i < n; i++) {
        int u, v; scanf("%d%d", &u, &v);
        G[u].push_back(v);
        G[v].push_back(u);
    }
    dfs(1, 0);
    solve(1, n);
    for(int i = 1; i <= n; i++) printf("%d
", ans[i]);
    return 0;
}

/*
*/