題意

給定一個\(1\)為根的樹，每個點有\(c,w\)兩個屬性，你需要從某個點\(u\)子樹裏選擇\(k\)個點，滿足選出來的點\(\sum_{i=1}^k w(i)\leq m\)，最大化\(k\times c(u)\)

題解

可以啟發式合並\(splay\)來做，\(\text{dfs}\)每個點，每次和兒子的\(splay\)合並，就得到了一個維護這個點子樹的平衡樹。再用這個點的\(w\)（題目中的領導力）乘以子樹中最多選多少結點滿足\(c\)（薪水）和\(\leq m\)

肯定貪心選，小的先選

然後這玩意我一開始腦抽寫了二分\(+kth\)的兩個\(\log\)做法，會\(\text{TLE}\)

int kth(int u, int k) {
    while(1) {
        if(sz[ch[u][0]] >= k) u = ch[u][0];
        else {
            k -= sz[ch[u][0]] + 1;
            if(k <= 0) break ;
            u = ch[u][1];
        }
    }
    splay(u);
    return u;
}

ll qsum(int &u, int k) {
    u = kth(u, k);
    return sum[ch[u][0]] + w[u];
}
ll query(int &u) {
    int l = 1, r = sz[u], ans = 0;
    while(l <= r) {
        int mid = (l + r) >> 1;
        if(qsum(u, mid) <= m) l = (ans = mid) + 1;
        else r = mid - 1;
    }
    return ans;
}
 

後來發現直接遞歸找就行了，十分簡單，從左子樹往自己往右子樹依次判能不能選進去

最後的代碼就是這樣了qwq：

#include <algorithm>
#include <cstdio>
#include <vector>
using namespace std;

typedef long long ll;

const int N = 2e5 + 10;

int n, m, w[N], c[N];
int fa[N], sz[N], ch[N][2];
vector<int> G[N];
ll ans, sum[N];

int dir(int u) { return ch[fa[u]][1] == u; }
void upd(int u) {
    sum[u] = sum[ch[u][0]] + sum[ch[u][1]] + w[u];
    sz[u] = sz[ch[u][0]] + sz[ch[u][1]] + 1;
}
void rotate(int u) {
    int d = dir(u), f = fa[u];
    if(fa[u] = fa[f]) ch[fa[u]][dir(f)] = u;
    if(ch[f][d] = ch[u][d ^ 1]) fa[ch[f][d]] = f;
    fa[ch[u][d ^ 1] = f] = u;
    upd(f); upd(u);
}
void ins(int &rt, int u, int f = 0) {
    if(!rt) {
        rt = u; fa[u] = f;
        sz[u] = 1; sum[u] = w[u];
        ch[u][0] = ch[u][1] = 0;
        return ;
    }
    ins(ch[rt][w[u] > w[rt]], u, rt);
    upd(rt);
}
void splay(int u) {
    for(; fa[u]; rotate(u)) if(fa[fa[u]])
        rotate(dir(u) == dir(fa[u]) ? fa[u] : u);
}
ll query(int u, ll s) {
    if(!u || !s) return 0;
    if(sum[u] <= s) return sz[u];
    if(sum[ch[u][0]] >= s) return query(ch[u][0], s);
    if(sum[ch[u][0]] + w[u] == s) return sz[ch[u][0]] + 1;
    if(sum[ch[u][0]] + w[u] > s) return sz[ch[u][0]];
    return sz[ch[u][0]] + 1 + query(ch[u][1], s - sum[ch[u][0]] - w[u]);
}

int a[N], l;
void get(int u) {
    if(u) {
        get(ch[u][0]);
        a[++ l] = u;
        get(ch[u][1]);
    }
}
int merge(int u, int v) {
    if(sz[u] < sz[v]) swap(u, v);
    l = 0; get(v); int rt = u;
    for(int i = 1; i <= l; i ++) {
        ins(rt, a[i]);
        splay(rt = a[i]);
    }
    return rt;
}

int dfs(int u) {
    int rt = u; sz[u] = 1; sum[u] = w[u];
    fa[u] = ch[u][0] = ch[u][1] = 0;
    for(int i = 0; i < G[u].size(); i ++)
        rt = merge(rt, dfs(G[u][i]));
    ans = max(ans, c[u] * query(rt));
    return rt;
}

int main() {
    scanf("%d%d", &n, &m);
    for(int f, i = 1; i <= n; i ++) {
        scanf("%d%d%d", &f, &w[i], &c[i]);
        G[f].push_back(i);
    }
    dfs(1);
    printf("%lld\n", ans);
    return 0;
}
 

「BZOJ 2809」「APIO 2012」Dispatching「啟發式合並」