DAG上的DP

DAG是有向无环图
而DAG的dp主要是利用一些问题的二元关系构造DAG图建模，转化成在图上求最长/短路的问题

https://www.luogu.com.cn/problem/UVA437

Code

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#include <bits/stdc++.h>
using namespace std;
// typedef long long ll;
#define int long long
typedef unsigned long long ull;
#define rep(i, a, n) for (int i = a; i <= n; i++)
#define req(i, n, a) for (int i = n; i >= a; i--)
typedef pair<int, int> pa;
int mod        = 998244353;
const int MAXN = 0x3f3f3f3f;
const int maxn = 2e5 + 10;
int n, x[40], y[40], z[40];
int dp[40][4];
//dp[i][j]是在以第i个砖为底，j式的所能堆塔的最大高度
int dfs(int k, int f) {
    //dp最大的dp[k][f]
    if (dp[k][f]) return dp[k][f];
    int data1, data2;
    if (f == 1) data1 = y[k], data2 = z[k];
    if (f == 2) data1 = x[k], data2 = z[k];
    if (f == 3) data1 = x[k], data2 = y[k];
    for (int i = 1; i <= n; i++) {
        if ((data1 > y[i] && data2 > z[i]) || (data2 > y[i] && data1 > z[i])) {
            dp[k][f] = max(dp[k][f], dfs(i, 1) + x[i]);
        }
        if ((data1 > x[i] && data2 > z[i]) || (data2 > x[i] && data1 > z[i])) {
            dp[k][f] = max(dp[k][f], dfs(i, 2) + y[i]);
        }
        if ((data1 > x[i] && data2 > y[i]) || (data2 > x[i] && data1 > y[i])) {
            dp[k][f] = max(dp[k][f], dfs(i, 3) + z[i]);
        }
    }
    return dp[k][f];
}

void solve() {
    int cnt = 0;
    while (++cnt) {
        cin >> n;
        if (!n) break;
        for (int i = 1; i <= n; i++) {
            cin >> x[i] >> y[i] >> z[i];
        }
        int high = 0;
        memset(dp, 0, sizeof(dp));
        for (int i = 1; i <= n; i++) {
            high = max(high, dfs(i, 1) + x[i]);
            high = max(high, dfs(i, 2) + y[i]);
            high = max(high, dfs(i, 3) + z[i]);
        }
        cout << "Case " << cnt << ": maximum height = " << high << '\n';
    }
}

signed main() {
    ios::sync_with_stdio(0);
    cin.tie(0), cout.tie(0);
    int TT = 1;
    // cin >> T;
    while (TT--) {
        solve();
    }
#ifdef local
    std::cout << endl, system("pause");
#endif
    return 0;
}
// cout << fixed << setprecision(2) <<
// priority_queue<pa, vector<pa>, greater<pa>>qx;
// priority_queue<int, vector<int>, less<int>>qd;
// struct node {
//     int id, f;
//     bool operator<(const node& x) const { return id > x.id; }
// };priority_queue<node> q;