Java 解法, 执行用时: 1440ms, 内存消耗: 37928K, 提交时间: 2022-04-26 15:55:35

import java.util.Scanner;
import static java.lang.Math.max;
import static java.lang.Math.min;

public class Main {
    public static void main(String[] args) {
        int n, a, b, c;
        int o = 0;
        int t = 60000;
        int[] p = new int[30010];
        Scanner in = new Scanner(System.in);
        n = in.nextInt();
        p[0] = 0;
        for (int i = 1; i <= n; i++) {
            a = in.nextInt();
            b = in.nextInt();
            c = in.nextInt();
            o += max(a, max(b, c));
            for (int j = o; j >= 0; j--) {
                if (b != 0) p[j] += b;
                else p[j] = 60000;
                if (a != 0 && j >= a) p[j] = min(p[j], p[j - a]);
                if (c != 0 && j >= c) p[j] = min(p[j], p[j - c] + c);
            }
        }
        for (int i = 0; i <= o; i++) t = min(t, max(i, p[i]));
        System.out.println(t);
    }
}

C++(clang++ 11.0.1) 解法, 执行用时: 358ms, 内存消耗: 564K, 提交时间: 2023-08-07 13:05:31

#include<iostream>
#include<algorithm>

using namespace std;

const int N = 30010, INF = 0x3f3f3f3f;
int t1, t2, t3, up;
int n;
int f[N];
int main()
{
    cin>>n;
    for(int i = 1; i <= n; i ++)
    {
        scanf("%d%d%d",&t1, &t2, &t3);
        up += max(t1, max(t2, t3));
        for(int j = up; j >= 0; j --)
        {
            int tem = INF;
            if(t1 && j - t1 >= 0)tem = min(tem, f[j - t1]);
            if(t2)tem = min(tem, f[j] + t2);
            if(t3 && j - t3 >= 0)tem = min(tem, f[j - t3] + t3);
            f[j] = tem;
        }
    }
    int res = INF;
    for(int i = 0; i <= up; i ++)res = min(res, max(i, f[i]));
    cout<<res<<endl;
    return 0;
}