zoukankan      html  css  js  c++  java
  • HDU 3449

    这个还没AC 目前TLE  滚动数组0.0

    Consumer

    Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/65536 K (Java/Others)
    Total Submission(s): 1023    Accepted Submission(s): 525


    Problem Description
    FJ is going to do some shopping, and before that, he needs some boxes to carry the different kinds of stuff he is going to buy. Each box is assigned to carry some specific kinds of stuff (that is to say, if he is going to buy one of these stuff, he has to buy the box beforehand). Each kind of stuff has its own value. Now FJ only has an amount of W dollars for shopping, he intends to get the highest value with the money.
     
    Input
    The first line will contain two integers, n (the number of boxes 1 <= n <= 50), w (the amount of money FJ has, 1 <= w <= 100000) Then n lines follow. Each line contains the following number pi (the price of the ith box 1<=pi<=1000), mi (1<=mi<=10 the number goods ith box can carry), and mi pairs of numbers, the price cj (1<=cj<=100), the value vj(1<=vj<=1000000)
     
    Output
    For each test case, output the maximum value FJ can get
     
    Sample Input
    3 800
    300 2 30 50 25 80
    600 1 50 130
    400 3 40 70 30 40 35 60
     
    Sample Output
    210
     
     1 #include <iostream>
     2 #include <cstdio>
     3 #include <algorithm>
     4 using namespace std;
     5 
     6 int main()
     7 {
     8     int j,k,n,m,V,p0,p,v;
     9     while(scanf("%d%d",&n,&V)!=EOF)
    10     {
    11         int d[100005]={0},dp[100005]={0};
    12         while(n--)
    13         {
    14             scanf("%d%d",&p0,&m);
    15             for(j=p0;j<=V;j++)
    16                 dp[j]=d[j-p0];
    17             while(m--)
    18             {
    19                 scanf("%d%d",&p,&v);
    20                 for(k=V-p;k>=p0;k--)
    21                     dp[k+p]=max(dp[k+p],dp[k]+v);
    22             }
    23             for(j=p0;j<=V;j++)
    24                 d[j]=max(d[j],dp[j]);
    25         }
    26         printf("%d
    ",d[V]);
    27     }
    28 
    29     return 0;
    30 }
  • 相关阅读:
    Pandas数据分析 (三)
    Pandas数据分析 (二)
    Pandas数据分析 (一)
    Django后台应用管理名称修改
    Jupyter Notebooks 配置
    Hadoop完全分布式搭建流程
    Java学习笔记(五)
    微信小程序学习笔记(一)
    redis windows版安装
    Linux计划任务
  • 原文地址:https://www.cnblogs.com/VOID-133/p/3641952.html
Copyright © 2011-2022 走看看