zoukankan      html  css  js  c++  java
  • Milking Time POJ

    Bessie is such a hard-working cow. In fact, she is so focused on maximizing her productivity that she decides to schedule her next N (1 ≤ N ≤ 1,000,000) hours (conveniently labeled 0..N-1) so that she produces as much milk as possible.

    Farmer John has a list of M (1 ≤ M ≤ 1,000) possibly overlapping intervals in which he is available for milking. Each interval i has a starting hour (0 ≤ starting_houriN), an ending hour (starting_houri < ending_houriN), and a corresponding efficiency (1 ≤ efficiencyi ≤ 1,000,000) which indicates how many gallons of milk that he can get out of Bessie in that interval. Farmer John starts and stops milking at the beginning of the starting hour and ending hour, respectively. When being milked, Bessie must be milked through an entire interval.

    Even Bessie has her limitations, though. After being milked during any interval, she must rest R (1 ≤ RN) hours before she can start milking again. Given Farmer Johns list of intervals, determine the maximum amount of milk that Bessie can produce in the N hours.

    Input

    * Line 1: Three space-separated integers: N, M, and R
    * Lines 2..M+1: Line i+1 describes FJ's ith milking interval withthree space-separated integers: starting_houri , ending_houri , and efficiencyi

    Output

    * Line 1: The maximum number of gallons of milk that Bessie can product in the N hours

    Sample Input

    12 4 2
    1 2 8
    10 12 19
    3 6 24
    7 10 31

    Sample Output

    43

    题解:排完序,状态就比较明显了,dp[ i ]表示第i个时间段能得到的最大牛奶数。
    因此 dp[ i ]=dp[ j ]+no[ i ](1<=j<i且满足条件After being milked during any interval,
    she must rest R (1 ≤ RN) hours before she can start milking again)
    #include<cstdio>
    #include<cstring>
    #include<iostream>
    #include<algorithm>
    using namespace std;
    
    const int maxn=1005;
    
    struct node{
        int l,r,sum;
        bool operator<(const node& i)const{
            return l < i.l;
        }
    }co[maxn];
    
    int n,m,re;
    int dp[maxn];
    
    void solve()
    {   int ans=0;
        for(int i=1;i<=m;i++){
            dp[i]=co[i].sum;
            for(int j=1;j<i;j++) if(co[i].l>=co[j].r+re) dp[i]=max(dp[i],dp[j]+co[i].sum);    
            ans=max(ans,dp[i]);
        }
        cout<<ans<<endl;
    }
  • 相关阅读:
    时间记录日志
    软件工程作业02
    个人学习进度(第二周)
    《大道至简》第二章读后感
    《大道至简》第一章读后感
    构建之法阅读笔记02
    构建之法阅读笔记01
    web开发
    Tomcat的安装与环境配置
    java-10异常处理动手动脑
  • 原文地址:https://www.cnblogs.com/zgglj-com/p/7287383.html
Copyright © 2011-2022 走看看