如题, 我的代码没有考虑一个房间不偷完的情况, 同时也没有考虑后来看题解才发现的$m$需要$-1$的"坑点"(汗

但是还是可以过这题, 代码如下, 请忽略菜鸡的码风以及将链式前向星转化成邻接表的行为以及其他槽点...(

#include <bits/stdc++.h>

#define max(a, b) ((a) > (b) ? (a) : (b))
#define min(a, b) ((a) < (b) ? (a) : (b))

using namespace std;

__attribute__((unused)) typedef long long ll;
__attribute__((unused)) typedef unsigned long long ull;

__attribute__((unused)) inline ll read() {
    ll x = 0, f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
        if (ch == '-')
            f = -1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
        x = (x << 1) + (x << 3) + (ch ^ 48);
        ch = getchar();
    }
    return x * f;
}

__attribute__((unused)) const int maxN = 1e2 + 4, maxM = 6e2 + 4, maxA = 14, maxB = 34, INF = 0x3f3f3f3f, MOD = 1e9 + 7;

int m;
struct node {
    int len = 0;
    int value = 0;
    int left = 0;
    int right = 0;

    node(int le = 0, int v = 0, int l = 0, int r = 0) : len(le), value(v), left(l), right(r) {}
};

int size = 0;
node a[maxN] = {};
int dp[maxN][maxM] = {};

void build(int t, int v) {
    int now = ++size;
    a[now].len = t;
    a[now].value = v;
    if (v)
        return;
    int tt = read(), vv = read();
    a[now].left = size + 1;
    build(tt, vv);
    tt = read(), vv = read();
    a[now].right = size + 1;
    build(tt, vv);
}

void dfs(int p) {
    const node &now = a[p];
    if (!now.left) {
        dp[p][2 * now.len + 5 * now.value] = now.value;
        return;
    }
    dfs(now.left);
    dfs(now.right);
    for (int i = m; i >= 2 * now.len; --i) {
        for (int j = 0; j <= i - 2 * now.len; ++j) {
            dp[p][i] = max(dp[p][i], dp[now.left][j] + dp[now.right][i - 2 * now.len - j]);
        }
    }
}

int main() {
//    debug();
    m = read();
    int t0 = read(), v0 = read();
    build(t0, v0);
    dfs(1);
    cout << dp[1][m];
}