POJ - 3616 Milking Time（DP+LIS）

POJ - 3616 Milking Time

Description

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

牛在n时间内产奶，农夫有m段时间可以挤奶，每一段时间有开始点strat，结束点end，和这段时间的挤奶量w。 每一次挤奶后，牛都必须休息r时间。问在最合理的挤奶安排下挤到的最大牛奶量是多少？

首先按照结束时间排序区间，然后按照第i个区间选择与否进行动态规划

#include<iostream>
#include<algorithm>
using namespace std;

const int N = 1010;
struct Node
{
    
	int l,r,val;
}a[N];
int dp[N];

bool cmp(Node A,Node B)
{
    
	return A.r<B.r;
}

int main()
{
    
	int n,m,r;cin>>n>>m>>r;
	for(int i=1;i<=m;i++)
	{
    
		cin>>a[i].l>>a[i].r>>a[i].val;
		a[i].r+=r;
	}
	sort(a+1,a+m+1,cmp);
	for(int i=1;i<=m;i++)
	{
    
		dp[i]=a[i].val;
		for(int j=1;j<i;j++)
			if(a[j].r<=a[i].l)
				dp[i]=max(dp[i],dp[j]+a[i].val);
	}
	int ans=0;
	for(int i=1;i<=m;i++)
		ans=max(ans,dp[i]);
	cout<<ans<<endl;
	return 0;
}