zoukankan      html  css  js  c++  java
  • HDU3466(01背包变种)

    Proud Merchants

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
    Total Submission(s): 4219    Accepted Submission(s): 1740


    Problem Description
    Recently, iSea went to an ancient country. For such a long time, it was the most wealthy and powerful kingdom in the world. As a result, the people in this country are still very proud even if their nation hasn’t been so wealthy any more.
    The merchants were the most typical, each of them only sold exactly one item, the price was Pi, but they would refuse to make a trade with you if your money were less than Qi, and iSea evaluated every item a value Vi.
    If he had M units of money, what’s the maximum value iSea could get?

     
    Input
    There are several test cases in the input.

    Each test case begin with two integers N, M (1 ≤ N ≤ 500, 1 ≤ M ≤ 5000), indicating the items’ number and the initial money.
    Then N lines follow, each line contains three numbers Pi, Qi and Vi (1 ≤ Pi ≤ Qi ≤ 100, 1 ≤ Vi ≤ 1000), their meaning is in the description.

    The input terminates by end of file marker.

     
    Output
    For each test case, output one integer, indicating maximum value iSea could get.

     
    Sample Input
    2 10
    10 15 10
     5 10 5
    3 10
    5 10 5
    3 5 6
    2 7 3
     
    Sample Output
    5
    11
    题意:有n个商品,买第i个需要Pi元,但手中的钱必须大于Qi元 ,所获得的价值为Vi,问给出M元,获得的最大价值为多少?
    思路: 每个商品按q-p排序才会得到正确结果
    #include"cstdio"
    #include"cstring"
    #include"algorithm"
    using namespace std;
    const int MAXN=1005;
    struct Node{
        int P,Q,v;
    }goods[505];
    int n,W;
    int dp[5005];
    int comp(Node a,Node b)
    {
        return a.Q-a.P < b.Q-b.P;
    }
    int main()
    {
        while(scanf("%d%d",&n,&W)!=EOF)
        {
            memset(dp,0,sizeof(dp));
            for(int i=0;i<n;i++)
            {
                scanf("%d%d%d",&goods[i].P,&goods[i].Q,&goods[i].v);
            }
            sort(goods,goods+n,comp);
            for(int i=0;i<n;i++)
            {
                for(int j=W;j>=goods[i].Q;j--)    dp[j]=max(dp[j],dp[j-goods[i].P]+goods[i].v);
            }
            printf("%d
    ",dp[W]);
        }
        
        return 0;
    }
     
  • 相关阅读:
    Dump 文件生成与分析
    打造支持apk下载和html5缓存的 IIS(配合一个超简单的android APP使用)具体解释
    GridView编辑删除操作
    google域名邮箱申请 gmail域名邮箱申请(企业应用套件)指南
    nvl,空时的推断和取值
    如何将图片保存至自定义分组
    Java实现 蓝桥杯VIP 算法训练 集合运算
    Java实现 蓝桥杯VIP 算法训练 瓷砖铺放
    Java实现 蓝桥杯VIP 算法训练 瓷砖铺放
    Java实现 蓝桥杯VIP 算法训练 集合运算
  • 原文地址:https://www.cnblogs.com/program-ccc/p/5187138.html
Copyright © 2011-2022 走看看