uva 10465 - Homer Simpson(贪心+完全背包)

最新推荐文章于 2015-06-26 23:04:00 发布

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动态规划（刘汝佳第九章）同时被 2 个专栏收录

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本文探讨了一个关于吃汉堡的优化问题，通过完全背包算法求解在有限时间内吃汉堡的最大数量，并考虑剩余时间的最小化。提供了一段AC代码实现，包括输入输出示例和解题思路。

1、http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1406

2、题目：给定一定时间，已知吃一种汉堡用时为m分钟，吃另一种汉堡用时为n分钟，求在保证剩余时间最少的情况下，吃的汉堡的最大数，如果时间有剩余，那么也要输出剩余的时间

题目可以看成是简单的完全背包，不过相当于两次，第一次是在有限时间t内，能装的最大时间，第二次是在有限时间内且剩余时间最少的情况下，能装的最大个数，

3、题目：

Problem C:	Homer Simpson
Time Limit: 3 seconds Memory Limit: 32 MB

Homer Simpson, a very smart guy, likes eating Krusty-burgers. It takes Homer m minutes to eat a Krusty- burger. However, there�s a new type of burger in Apu�s Kwik-e-Mart. Homer likes those too. It takes him n minutes to eat one of these burgers. Given t minutes, you have to find out the maximum number of burgers Homer can eat without wasting any time. If he must waste time, he can have beer.

Input

Input consists of several test cases. Each test case consists of three integers m, n, t (0 < m,n,t < 10000). Input is terminated by EOF.

Output

For each test case, print in a single line the maximum number of burgers Homer can eat without having beer. If homer must have beer, then also print the time he gets for drinking, separated by a single space. It is preferable that Homer drinks as little beer as possible.

Sample Input

3 5 54
3 5 55

Sample Output

18
17

Problem setter: Sadrul Habib Chowdhury
Solution author: Monirul Hasan (Tomal)

Time goes, you say? Ah no!
Alas, Time stays, we go.

4、AC代码：

#include<stdio.h>
#include<string.h>
#define N 10005
#include<algorithm>
using namespace std;
int dp[N];
int w[5];
int sum[N];
int main()
{
    int m,n,t;
    while(scanf("%d%d%d",&m,&n,&t)!=EOF)
    {
        w[1]=m;w[2]=n;
        memset(sum,0,sizeof(sum));
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=2;i++)
        {
            for(int j=0;j<=t;j++)
            {
                if(j>=w[i])
                {
                    if(dp[j]<dp[j-w[i]]+w[i])
                    {
                        dp[j]=dp[j-w[i]]+w[i];
                        sum[j]=sum[j-w[i]]+1;
                    }
                    else if(dp[j]==dp[j-w[i]]+w[i])//注意第一层的情况下，判断最优
                    sum[j]=max(sum[j],sum[j-w[i]]+1);
                }
            }
        }
        if(dp[t]==t)
        printf("%d\n",sum[t]);
        else
        printf("%d %d\n",sum[t],t-dp[t]);
    }
    return 0;
}