2021 Shanghai match-b-ranked DP

The main idea of the topic :

Here you are. $n$ A triad $(a_i,b_i,c_i)$. Each dimension should be selected separately $a,b,c(a+b+c=n)$ One makes the most value .

Topic ideas :

Start with greed :
In the case of two tuples , We just need to be right $b-a$ Choose greedily after sorting b individual $b_i$ The remaining choices $a_i$.

After digging out a subsequence of the sorted sequence, it still conforms to the nature of greed . Therefore, we can analyze the third dimension $dp(i,j)$ On behalf of i A triad , Yes $j$ individual $c_i$ Maximum value of .

Ruodi $i$ No choice , According to the nature , When $i\le b$ Greedy choice $b$, Reverse selection $a$.

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define pii pair<int,int>
#define pb push_back
#define mp make_pair
#define vi vector<int>
#define vll vector<ll>
#define fi first
#define se second
const int maxn = 5000 + 5;
const int mod = 1e9 + 7;
struct Node{
    
    int a , b , c;
    bool operator < (const Node & g){
    
        return b - a > g.b - g.a;
    }
}a[maxn];
ll dp[maxn][maxn];
int main()
{
    
    ios::sync_with_stdio(false);
    int n , x , y , z; cin >> n >> x >> y >> z;
    for (int i = 1 ; i <= n ; i++){
    
        cin >> a[i].a >> a[i].b >> a[i].c;
    }
    sort(a + 1 , a + 1 + n);
    for (int i = 0 ; i <= n ; i++)
        for (int j = 0 ; j <= n ; j++)
            dp[i][j] = -1e18;
    dp[0][0] = 0;
    for (int i = 1 ; i <= n ; i++){
    
        for (int j = 0 ; j <= i ; j++){
    
            if (j)
                dp[i][j] = dp[i - 1][j - 1] + a[i].c;
            int rest = i - j;
            if (rest <= y) dp[i][j] = max(dp[i][j] , dp[i - 1][j] + a[i].b);
            else dp[i][j] = max(dp[i][j] , dp[i - 1][j] + a[i].a);
        }
    }
    cout << dp[n][z] << endl;
    return 0;
}