2021上海市赛-B-排序后dp

题目大意:

给你$n$个三元组$(a_i,b_i,c_i)$.每个维度要分别选出$a,b,c(a+b+c=n)$个使得价值最大。

题目思路:

从贪心入手:
如果是二元组的情况,我们只需要对$b-a$排序后贪心选b个$b_i$再剩下的选$a_i$.

将排序后的序列挖去一个子序列后依然符合贪心的性质.所以我们可以对第三个维度进行$dp(i,j)$代表前i个三元组，选了$j$个$c_i$的最大价值.

若第$i$个不选,根据性质,当$i\le b$时贪心选$b$，反之选$a$.

#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;
}