zoukankan      html  css  js  c++  java
  • I love sneakers! 分组背包 二维的 值得思考

    Description

    After months of hard working, Iserlohn finally wins awesome amount of scholarship. As a great zealot of sneakers, he decides to spend all his money on them in a sneaker store.

    There are several brands of sneakers that Iserlohn wants to collect, such as Air Jordan and Nike Pro. And each brand has released various products. For the reason that Iserlohn is definitely a sneaker-mania, he desires to buy at least one product for each brand.
    Although the fixed price of each product has been labeled, Iserlohn sets values for each of them based on his own tendency. With handsome but limited money, he wants to maximize the total value of the shoes he is going to buy. Obviously, as a collector, he won’t buy the same product twice.
    Now, Iserlohn needs you to help him find the best solution of his problem, which means to maximize the total value of the products he can buy.
     

    Input

    Input contains multiple test cases. Each test case begins with three integers 1<=N<=100 representing the total number of products, 1 <= M<= 10000 the money Iserlohn gets, and 1<=K<=10 representing the sneaker brands. The following N lines each represents a product with three positive integers 1<=a<=k, b and c, 0<=b,c<100000, meaning the brand’s number it belongs, the labeled price, and the value of this product. Process to End Of File.
     

    Output

    For each test case, print an integer which is the maximum total value of the sneakers that Iserlohn purchases. Print "Impossible" if Iserlohn's demands can’t be satisfied.
     

    Sample Input

    5 10000 3 1 4 6 2 5 7 3 4 99 1 55 77 2 44 66
     

    Sample Output

    255
    ***************************************************************************************************************************
    分组背包,三重循环!
    ***************************************************************************************************************************
     1 #include<iostream>
     2 #include<string>
     3 #include<cstring>
     4 #include<cstdio>
     5 #include<algorithm>
     6 using namespace std;
     7 int dp[20][10011],i,j,k,st;
     8 int n,sum,m;
     9 struct node
    10 {
    11     int id;
    12     int price;
    13     int value;
    14 }e[1001];
    15 
    16 void pack()
    17 {
    18     memset(dp,0,sizeof(dp));
    19     for(i=1;i<=m;i++)
    20      for(j=1;j<=n;j++)
    21       for(k=sum;k>=0;k--)
    22         if(e[j].id==i&&e[j].price<=k)
    23           dp[i][k]=max(dp[i][k],max(dp[i][k-e[j].price]+e[j].value,dp[i-1][k-e[j].price]+e[j].value));
    24 }
    25 int main()
    26 {
    27    while(scanf("%d%d%d",&n,&sum,&m)!=EOF)
    28     {
    29        for(i=1;i<=n;i++)
    30        {
    31            scanf("%d %d %d",&e[i].id,&e[i].price,&e[i].value);
    32        }
    33        pack();
    34       if(dp[m][sum]==0)
    35         printf("Impossible
    ");
    36        else
    37         printf("%d
    ",dp[m][sum]);
    38     }
    39     return 0;
    40 }
    View Code
  • 相关阅读:
    C加加学习之路 1——开始
    哈夫曼树C++实现详解
    Linux常用命令
    Accp第二章:基础知识
    第一章Accp 8.0
    泛型集合
    深入C#数据类型
    初始wondows系统
    深入.NET框架
    二至十五章总结
  • 原文地址:https://www.cnblogs.com/sdau--codeants/p/3354483.html
Copyright © 2011-2022 走看看