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

The cow is in n Milk production in time , Farmers have m You can milk for a while , Every period of time has a starting point strat, The end point end, And the milking volume during this period w. After each milking , Cows must rest r Time . Ask how much milk you can get under the most reasonable milking arrangement ？

First, sort the interval according to the end time , Then according to the i Whether to choose or not to carry out dynamic planning

#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;
}