Question

我正在努力解决一个编程问题，为明天的比赛练习，我想也许这将是一个讨论如何接近它的好地方。问题是此网站上的第一个问题：http://www.cs.rit.edu/~icpc/questions/2010/Oswego_2010.pdf

本网站上的常见问题解答提到了算法和数据结构的概念，以及设计模式，所以我想问一下如何解决这个问题并非偏离主题。这是我到目前为止（不多）。我不明白如何解决这个问题。

public class Ape
{
    public void computeOutput(int weight, int[] capacities, int[] snackLosses)
    {
        //not sure what to do
    }

    public static void main(String [] args) throws FileNotFoundException
    {
        Ape ape = new Ape();
        File file = new File(args[0]);
        Scanner in = new Scanner(file);
        int totalWeight = in.nextInt();
        int n = in.nextInt();
        int[] capacities = new int[n];
        int[] snackLosses = new int[n];

        for (int i = 0; i < n; i++)
        {
            capacities[i] = in.nextInt();
            snackLosses[i] = in.nextInt();
        }

        ape.computeOutput(totalWeight, capacities, snackLosses);
    }
}

Answer 1

乍一看，这似乎是一个动态编程问题。

基本上，我们有一个函数f（N，K）=给予K的bannas和前N只猴子带来的bannas数量。

显然f（0，K）= 0且f（N，0）= 0

然后你所要做的就是找出f（n，k）的值。你应该通过最多两个案例来做到这一点：

猴子不接受bannana f（n，k）= f（n-1，k），因为猴子什么都不做，就像他不在那里一样
猴子取bannana f（n，k）= f（n-1，k - 力量）+力量 - 东西吃猴子

使用此逻辑填写表格，然后确定f（N，K），你就得到了答案。

Answer 2

这个问题可以简化为0/1背包，这本身就是众所周知的DP算法。但是这里没有价值，你有能力 - 小吃。

Answer 3

#include <iostream>
using namespace std;
main(){
int totalwight,numberapes;
//cout<<"Enter The total weight of bananas available to lug home : ";
cin>>totalwight;
//cout<<"Enter The number of apes : ";
cin>>numberapes;
int *capacity = new int [numberapes];
int *tcapacity = new int [numberapes];
int *snackloss = new int [numberapes];
int *pos = new int [numberapes];
int *tpos = new int [numberapes];
for (int i=0 ; i<numberapes ; i++){
     pos[i]=i+1;
     tpos[i]=0;
}

for (int i=0 ; i<numberapes ; i++){
   // cout<<"Enter How many monkey # "<<i+1<<" carrying capacity : ";
    cin>>capacity[i];
    tcapacity[i]=capacity[i];
    //cout<<"Enter How many snack loss of monkey # "<<i+1<<" : ";
    cin>>snackloss[i];
}
int *arr = new int [numberapes];
for (int i=0 ; i<numberapes ; i++)
    arr[i] = capacity[i] - snackloss[i];
    int temp;
for (int i=0 ; i<numberapes ; i++)
    for (int j=0 ; j<i ; j++)
        if (arr[i] > arr[j]){
            temp = arr[i];
            arr[i] = arr[j];
            arr[j] = temp;
            temp = tcapacity[i];
            tcapacity[i] = tcapacity[j];
            tcapacity[j] = temp;
            temp = pos[i];
            pos[i] = pos[j];
            pos[j] = temp;
        }
int *tarr = new int [numberapes];
int j=0;
for (int i=0 ; i<numberapes ; i++)
    tarr[i]=0;
for (int i=0 ; i<numberapes ; i++){
      if (arr[i] <= totalwight && tcapacity[i] <= totalwight){
            totalwight -= tcapacity[i];
            tpos[j] = pos[i];
            j++;
      }
}
for (int i=0 ; i<numberapes ; i++)
        for (int j=0 ; j<numberapes ; j++)
            if (tpos[j] > tpos[i] && tpos[i]!=0 && tpos[j]!=0){
                temp = tpos[i];
                tpos[i] = tpos[j];
                tpos[j] = temp;
            }
int t1=1,t2=0;
while (t1<=numberapes){
    if (t1 == tpos[t2]){
        cout<<"1 ";
        t2++;
    }
    else
        cout<<"0 ";
    t1++;
}

}

Answer 4

请查看我的答案：

using System;
using System.IO;
using System.Linq;
using System.Collections.Generic;

namespace monkey
{
    class Program
    {

        static int getTotalCapacity(int[] ki, int[] indexes)
        {
            int result = 0;
            foreach (int i in indexes)
            {
                result += ki[i];
            }
            return result;

        }

        static int[] computeOutput(int k, int n, int[] ki, int[] wi)
        {
            // #1: Sort the input by ki - wi
            Dictionary<int, int> vi = new Dictionary<int, int>();
            for (int i = 0; i < n; i++)
            {
                vi.Add(i, ki[i] - wi[i]);
            }
            Dictionary<int, int> sorted_vi = (from entry in vi orderby entry.Value descending select entry).ToDictionary(x => x.Key, x => x.Value);

            // #2: Take elements in sequence in order to fulfill the minimum total k
            Dictionary<int, int> result = new Dictionary<int, int>();
            foreach (KeyValuePair<int, int> kv in sorted_vi)
            {
                if (ki[kv.Key] + getTotalCapacity(ki, result.Select(x=>x.Key).ToArray()) <= k)
                {
                    result.Add(kv.Key,kv.Value);
                }
            }

            // #3 Transform to output format
            int[] output = new int[n];
            foreach(KeyValuePair<int,int> kv in result){
                output[kv.Key] = 1;
            }

            return output;
        }

        static void Main(string[] args)
        {
            if (args.Length != 1)
            {
                Console.WriteLine("Usage: <prog>.exe <filepath>");
                return;
            }
            try
            {
                int k;
                int n;
                int[] ki;
                int[] wi;

                // Reading file input
                using (StreamReader fs = new StreamReader(args[0]))
                {
                    k = Convert.ToInt32(fs.ReadLine());
                    n = Convert.ToInt32(fs.ReadLine());
                    ki = new int[n];
                    wi = new int[n];
                    for (int i = 0; i < n; i++)
                    {
                        string[] line = fs.ReadLine().Split(new string[] { " " }, StringSplitOptions.RemoveEmptyEntries);
                        ki[i] = Convert.ToInt32(line[0]);
                        wi[i] = Convert.ToInt32(line[1]);
                    }
                }

                int[] output = computeOutput(k, n, ki, wi);
                Console.WriteLine(string.Join(" ", output));

            }
            catch (Exception ex)
            {
                Console.WriteLine(ex.Message);
                return;
            }
        }
    }
}

ACM编程问题

4 个答案: