zoukankan      html  css  js  c++  java
  • HDU3466——背包DP——Proud Merchants

    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
     大意:考虑到了要排序..按照q-p进行排序,从小到大,q-p即下面一个物品的最少的所需要的钱,贪心的思想,然后01背包
    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    using namespace std;
    int dp[5500];
    int n,m;
    struct edge{
        int p;
        int q;
        int v;
    }a[5500];
    bool cmp(edge i, edge j){
        return i.q - i.p <j.q - j.p;
    }
    int main()
    {
        while(~scanf("%d%d",&n,&m)){
            memset(dp,0,sizeof(dp));
            for(int i = 1; i <= n ; i++)
                scanf("%d%d%d",&a[i].p,&a[i].q,&a[i].v);
            sort(a+1,a+n+1,cmp);
            for(int i = 1 ; i <= n ; i++){
                for(int j = m; j >= a[i].q ;j--){
                    dp[j] = max(dp[j],dp[j-a[i].p]+a[i].v);
                }
            }
            printf("%d
    ",dp[m]);
        }
        return 0;
    }
    View Code
  • 相关阅读:
    ASP.NET Web API模型验证以及异常处理方式
    Javascript基础恶补
    求一个集合的集合下所有集合元素求值
    C#创建唯一的订单号, 考虑时间因素
    git的几十个基本面
    报错:ASP.NET Web API中找不到与请求匹配的HTTP资源
    使用RAML描述API文档信息的一些用法整理
    Postman测试Web API
    Javascript中的Prototype到底是啥
    AngularJS和DataModel
  • 原文地址:https://www.cnblogs.com/zero-begin/p/4455493.html
Copyright © 2011-2022 走看看