AcWing 4548. 猴子和香蕉

设 $f_i$ 表示对于前 $i$ 组，必定选第 $i$ 个积木的最大可能高度，$f_i=max{f}_k+z_i$ 然后一开始往数组中扔进 $6$ 个不同面的积木，然后拍个序去除后效性即可（虽然本来就没有后效性）。

注意：位于下面的积木的长和宽必须严格大于位于上面的积木的长和宽。每种积木的供应数量无限多。自由决定积木的哪个面作为底面，以及积木的具体摆放朝向。

// #define FILE_INPUT
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

#define rep(i, a, b) for (int i = a, END##i = b; i <= END##i; i++)
#define per(i, a, b) for (int i = a, END##i = b; i >= END##i; i--)

void Init();
void Solve();

signed main() {
    cin.sync_with_stdio(0);
    cin.tie(0), cout.tie(0);

    #ifdef FILE_INPUT
        freopen("input.in", "r", stdin);
    #endif

    int T = 1;
    // cin >> T;
    while (T--) {
        Init();
        Solve();
    }
    return 0;
}

using LL = long long;
using ULL = unsigned long long;

const int Mod = 1e9 + 7;
const int Inf = 0x3f3f3f3f;
const LL InfLL = 0x3f3f3f3f3f3f3f3f;

const int N = 1e6 + 10;
int n, f[N], cnt;
struct Node {
    int x, y, z;
    Node() {};
    Node(int x, int y, int z) : x(x), y(y), z(z) {};
    bool operator< (const Node& b) const {
        return x == b.x ? y < b.y : x < b.x;
    }
}a[N];

void Init() {
}

int cases;

void Solve() {
    while (cin >> n, n) {
        cnt = 0;
        rep(i, 1, n) {
            int x, y, z; cin >> x >> y >> z;
            a[++cnt] = Node(x, y, z);
            a[++cnt] = Node(x, z, y);
            a[++cnt] = Node(y, x, z);
            a[++cnt] = Node(y, z, x);
            a[++cnt] = Node(z, x, y);
            a[++cnt] = Node(z, y, x);
        }
        sort(a + 1, a + cnt + 1);
        int ans = 0;
        rep(i, 1, cnt) {
            f[i] = a[i].z;
            rep(j, 1, i - 1) {
                if (a[i].x == a[j].x || a[i].y <= a[j].y) continue;
                f[i] = max(f[i], f[j] + a[i].z);
            }
            ans = max(ans, f[i]);
        }
        cout << "Case " << ++cases << ": maximum height = " << ans << "\n";
    }
}