POJ 3616 Milking Time (DP)

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_hour_i ≤ N), an ending hour (starting_hour_i < ending_hour_i ≤ N), and a corresponding efficiency (1 ≤ efficiency_i ≤ 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_hour_i , ending_hour_i , and efficiency_i

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

题意：你有一头奶牛，你能够在一定的时间里挤奶。而且挤奶量也不同，每次挤奶要休息r时间，问你最大可以挤多少奶。

水题一道，看完就知道一定是对时间段dp，然后就是两个for的事了。只要前面能满足条件的状态就可以转移过来，然后取最大，不过要先排序。

状态设定：dp[i]表示从开始取，到满足取第i段的最优值。

#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;
struct eg
{
	int x,y,w;
}e[1005];
bool cmp(eg a,eg b)
{
	if(a.y==b.y)
		return a.x<b.x;
	return a.y<b.y;
}
int main()
{
	int n,m,r,dp[1005],ans;
	scanf("%d%d%d",&n,&m,&r);
	for(int i=0;i<m;i++)
		scanf("%d%d%d",&e[i].x,&e[i].y,&e[i].w);
	sort(e,e+m,cmp);
	ans=0;
	for(int i=0;i<m;i++)
	{
	    dp[i]=e[i].w;
		for(int j=0;j<i;j++)
		{
			if(e[i].x>=e[j].y+r)
			    dp[i]=max(dp[i],dp[j]+e[i].w);
		}
		ans=max(ans,dp[i]);
	}
	printf("%d
",ans);
	return 0;
}