本文主要是介绍D - Milking Time，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
Milking Time
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.
* 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
* Line 1: The maximum number of gallons of milk that Bessie can product in the N hours
12 4 2 1 2 8 10 12 19 3 6 24 7 10 31
43
这是刚开始写的代码，超时，唉，我就知道不会这么简单~~
#include<iostream>
#include<algorithm>
using namespace std;struct node
{int start,end,eff;int time;
}s[1001];int cmp(node a,node b)
{return a.start<b.start;
}int main()
{int n,m,r;cin>>n>>m>>r;for(int i=0;i<m;i++){cin>>s[i].start>>s[i].end>>s[i].eff;s[i].time=s[i].end-s[i].start;}sort(s,s+m,cmp);int max=0;for(int i=0;i<n;i++){int sum=s[i].eff;int end1=s[i].end;for(int j=i+1;j<n;j++){if(end1+r<=s[j].start){sum+=s[j].eff;end1=s[j].end;}}if(sum>max)max=sum;}cout<<max<<endl;return 0;
}正确的的代码：
#include<iostream>
#include<algorithm>
using namespace std;int dp[1001]; //dp[i]表示挤完第i个时间段的奶的最大挤奶量struct node
{int start,end,eff;int time;
}s[1001];int cmp(node a,node b)
{return a.start<b.start;
}int main()
{int n,m,r;cin>>n>>m>>r;for(int i=0;i<m;i++){cin>>s[i].start>>s[i].end>>s[i].eff;s[i].end+=r;}sort(s,s+m,cmp);for(int i=0;i<m;i++){dp[i]=s[i].eff;for(int j=0;j<i;j++){if(s[j].end<=s[i].start)dp[i]=max(dp[i],dp[j]+s[i].eff);}}sort(dp,dp+m); //若第1、2个时间段都只能挤一次，则dp[1]=s[1].eff，dp[2]=s[2].effcout<<dp[m-1]<<endl; //此时要判断dp[1]dp[2]即s[1].eff，s[2].eff的大小return 0;
}
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