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  • hdu1712 分组背包

    ACboy needs your help

    Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 5964    Accepted Submission(s): 3251


    Problem Description
    ACboy has N courses this term, and he plans to spend at most M days on study.Of course,the profit he will gain from different course depending on the days he spend on it.How to arrange the M days for the N courses to maximize the profit?
     
    Input
    The input consists of multiple data sets. A data set starts with a line containing two positive integers N and M, N is the number of courses, M is the days ACboy has.
    Next follow a matrix A[i][j], (1<=i<=N<=100,1<=j<=M<=100).A[i][j] indicates if ACboy spend j days on ith course he will get profit of value A[i][j].
    N = 0 and M = 0 ends the input.
     
    Output
    For each data set, your program should output a line which contains the number of the max profit ACboy will gain.
     
    Sample Input
    2 2 1 2 1 3 2 2 2 1 2 1 2 3 3 2 1 3 2 1 0 0
    Sample Output
    3 4 6
    题目大意:ACboy要开始选课了,上一门课能够获得的收益和他上这门课的时间是有关的,然后给你若干门课,让你帮他进行选课,
    每一门课自然是只能选择一个课程时长,问你如何选择,才能使ACboy获得的受益最大。
    思路分析:这是自己做的第一道分组背包的题目,抽象一下,背包容量相当于ACboy拥有的时间,然后给你若干组物品(课程),每一组
    物品里面有着若干个物品(课程时间),一组里面的物品只能选择一个,这么一抽象,标准的分组背包问题,相当于若干组物品的01背包
    问题,但是要注意,v的循环一定要放在组内物品循环之外,这样才能保证一个组的物品只用了一个,另外,内层循环要有一个判断,判断是
    否满足v-cost>=0,否则会出现数组越界,报错。
    代码:#include <iostream>
    #include <stack>
    #include <cstdio>
    #include <cstring>
    #include <cstring>
    #include <algorithm>
    #include <cmath>
    using namespace std;
    const int maxn=105;
    int a[maxn][maxn];
    int f[maxn];
    int main()
    {
       int n,m;
       while(scanf("%d%d",&n,&m)&&(n||m))
       {
           memset(f,0,sizeof(f));
           for(int i=1;i<=n;i++)
            for(int j=1;j<=m;j++)
            scanf("%d",&a[i][j]);
           for(int i=1;i<=n;i++)
           {
               for(int j=m;j>=0;j--)
               {
                   for(int k=1;k<=m;k++)
                   {
                      if(j-k>=0)
                       f[j]=max(f[j],f[j-k]+a[i][k]);
                   }
               }
           }
         cout<<f[m]<<endl;
       }
    }
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  • 原文地址:https://www.cnblogs.com/xuejianye/p/5426377.html
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