思路

可并堆思路：自顶向上做大根堆，然后弹出堆顶直到整个堆的和小于 m。（性质：父亲节点不会选择子树不包含的忍者）。

线段树合并思路：没看懂，不写了。

代码

// P1552.cpp
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <vector>
#define ll long long
using namespace std;
const int MAX_N = 100100;
struct node
{
    int lson, rson;
    ll dist;
} nodes[MAX_N];
ll n, m, li[MAX_N], ci[MAX_N], master, ans = -2e9;
ll sum[MAX_N], siz[MAX_N], pt[MAX_N];
vector<int> G[MAX_N];
int merge(int x, int y)
{
    if (!x || !y)
        return x + y;
    if (ci[x] < ci[y])
        swap(x, y);
    nodes[x].rson = merge(nodes[x].rson, y);
    if (nodes[nodes[x].lson].dist < nodes[nodes[x].rson].dist)
        swap(nodes[x].lson, nodes[x].rson);
    nodes[x].dist = nodes[nodes[x].rson].dist + 1;
    return x;   
}
void dfs(int u)
{
    sum[u] = ci[u], siz[u] = 1, pt[u] = u;
    int siz_ = G[u].size();
    for (int i = 0; i < siz_; i++)
    {
        int to = G[u][i];
        dfs(to);
        sum[u] += sum[to], siz[u] += siz[to];
        pt[u] = merge(pt[u], pt[to]);
    }
    while (sum[u] > m && siz[u])
    {
        sum[u] -= ci[pt[u]];
        pt[u] = merge(nodes[pt[u]].rson, nodes[pt[u]].lson);
        siz[u]--;
    }
    ans = max(ans, li[u] * siz[u]);
}
int main()
{
    scanf("%lld%lld", &n, &m);
    for (int i = 1; i <= n; i++)
    {
        ll fa;
        scanf("%lld%lld%lld", &fa, &ci[i], &li[i]);
        if (fa == 0)
            master = i;
        else
            G[fa].push_back(i);
    }
    dfs(master);
    printf("%lld", ans);
    return 0;
}

我之前把数据结构搞复杂了，看了题解发现好简洁。还是要做减法啊。

神仙思路

有一个“显然”的定理：如果满足\(n\&k==k\)，那么\(C^k_n\)为奇数。具体感性证明：litble 的证明。所以，直接上代码。

代码

// P3773.cpp
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int MAX_N = 233393, mod = 1000000007;
int n, arr[MAX_N], bucket[MAX_N], f[MAX_N], ans;
int getMod(int num) { return num >= mod ? num - mod : num; }
int main()
{
    scanf("%d", &n);
    for (int i = 1; i <= n; i++)
        scanf("%d", &arr[i]), bucket[arr[i]] = i;
    for (int i = n; i >= 1; i--)
    {
        f[i] = 1;
        for (int j = (arr[i] & (arr[i] - 1)); j; j = arr[i] & (j - 1))
            if (bucket[j] > i)
                f[i] = getMod(f[i] + f[bucket[j]]);
        ans = getMod(ans + f[i]);
    }
    ans = (ans - n + mod) % mod;
    printf("%d", ans);
    return 0;
}