Solution

想存一下左偏树的模板。

我们发现容量上限是不变的，那么就可以贪心地依次选取每个点子树中的最大值删去。

就可以啦。

Code

#include <cstdio>
#include <iostream>
using namespace std;
typedef long long ll;

const int N = 1e5 + 5;

int head[N], dot[N << 1], nxt[N << 1], cnt, m;
ll ans;

struct LT {
    int f[N], son[N][2], val[N], dis[N], siz[N];
    ll s[N], l[N];

    void init(const int n) {
        for(int i = 1; i <= n; ++ i) {
            son[i][0] = son[i][1] = dis[i] = s[i] = 0;
            f[i] = i; siz[i] = 1;
        }
    }

    int Find(const int x) {return x == f[x] ? x : f[x] = Find(f[x]);}

    bool cmp(const int x, const int y) {return Find(x) == Find(y);}

    int unite(int x, int y) {
        if(! x || ! y) return x + y;
        if(val[x] < val[y]) swap(x, y);
        son[x][1] = unite(son[x][1], y); f[son[x][1]] = x;
        if(dis[son[x][0]] < dis[son[x][1]]) swap(son[x][0], son[x][1]);
        dis[x] = dis[son[x][1]] + 1;
        return x;
    }

    int del(const int x) {
        int l = son[x][0], r = son[x][1];
        son[x][0] = son[x][1] = dis[x] = 0;
        f[l] = l; f[r] = r;
        return unite(l, r);
    }

    void solve(const int x, const int fa) {
        for(int i = head[x]; i; i = nxt[i]) {
            int v = dot[i];
            if(v == fa) continue;
            solve(v, x);
            siz[x] += siz[v]; s[x] += s[v];
            f[x] = unite(f[x], f[v]);
        }
        while(s[x] > m) {
            s[x] -= val[f[x]]; -- siz[x];
            f[x] = unite(son[f[x]][0], son[f[x]][1]);
        }
        ans = max(ans, l[x] * siz[x]);
    }
}T;

void addEdge(const int u, const int v) {
    dot[++ cnt] = v; nxt[cnt] = head[u]; head[u] = cnt;
}

int read() {
    int x = 0, f = 1; char s;
    while((s = getchar()) < ‘0‘ || s > ‘9‘) if(s == ‘-‘) f = -1;
    while(s >= ‘0‘ && s <= ‘9‘) {x = (x << 1) + (x << 3) + (s ^ 48); s = getchar();}
    return x * f;
}

int main() {
    int x, n, root;
    n = read(), m = read();
    T.init(n);
    for(int i = 1; i <= n; ++ i) {
        x = read(), T.s[i] = T.val[i] = read(), T.l[i] = read();
        if(x) addEdge(x, i);
        else root = i;
    }
    T.solve(root, 0);
    printf("%lld\n", ans);
    return 0;
}

【APIO 2012】dispatching