Milking Time-白红宇

Milking Time

阅读量：206 次

发布时间：2019-02-28

本文共 2463 字，大约阅读时间需要 8 分钟。

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_houri ≤ N), an ending hour (starting_houri < ending_houri ≤ N), 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 ≤ R ≤ N) 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

题目大意：一头奶牛现在要工作n小时，在这n小时里有m个任务。奶牛每完成一个任务后要休息r小时。给出这m个任务的开始时间、结束时间和得到的奶量。求在这m个任务中如何选择能得到最大的产奶量。

题目分析：

状态表示：f[i] //表示1-i小时的最大产奶量

预处理：因为奶牛在完成了一个任务的时候会休息r小时，我们可以将这r小时加到区间右端点上。(即:a[i].r+=rest)
注意：在状态计算的时候也要往后算到f[n+rest]。
然后将这m个区间按照右端点进行排序(按左端点进行排序好像也可以，但那样计算方法也会不一样)。

状态计算：有两个状态
1）当第i小时没有完成某个任务产奶时，f[i]=f[i-1] //1一i小时的最大产奶量等于1一i-1小时的产奶量
2）当第i个小时完成了某些任务产奶时，即i==a[k].r（1<=k<=m），有两个选择(选择该任务和不选该任务),取最大值即可。f[i]=max(f[i-1],f[a[k].l]+a[k++].c)

代码如下：

#include 
   
    #include 
    
     #include 
     
      #include 
      
       #include 
       
        #include 
         #include 
         
          #include 
          
           #include 
           
            #define LL long longconst int N=1e3+5,M=1e6+5;using namespace std;struct Node{ int l,r,c; //每个任务的开始时间、结束时间以及产奶量 bool operator< (const Node &a) const { return r
            
             >n>>m>>rest; for(int i=1;i<=m;i++) { cin>>a[i].l>>a[i].r>>a[i].c; a[i].r+=rest; //预处理 } sort(a+1,a+1+m); //按右端点进行排序 int ans=0,k=1; for(int i=1;i<=n+rest;i++) { f[i]=f[i-1]; //第一种情况没有任何限制条件 //因为右端点从小到大排过序，因此计算后直接让k++，取下一个即可 while(a[k].r==i) f[i]=max(f[i],f[a[k].l]+a[k++].c); } //以i为右端点的区间可能有多个，要用while cout<
             
              <

转载地址：http://mptn.baihongyu.com/

你可能感兴趣的文章

Mysql_Postgresql中_geometry数据操作_st_astext_GeomFromEWKT函数_在java中转换geometry的16进制数据---PostgreSQL工作笔记007