Milking Time -最長遞增子序列變形題

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 Nhours

Sample Input

12 4 2
1 2 8
10 12 19
3 6 24
7 10 31

Sample Output

43

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<queue>
#include<map>
#include<set>
#define inf 0x3f3f3f3f
#define ll long long
using namespace std;
struct node{
	int x,y,step;
}pp[1000009];
bool cmp(node a,node b)
{
	if(a.y == b.y)
	{
		return a.x < b.x;
	}
	return a.y < b.y;
}//快排利用貪心的思想先貪心最小的結束時間
int dp[10000009];
int main()
{
	int x,m,r;
	scanf("%d%d%d",&x,&m,&r);
	for(int i = 1;i <= m;++i)
	{
		scanf("%d%d%d",&pp[i].x,&pp[i].y,&pp[i].step);
	}
	sort(pp+1,pp+m+1,cmp);
	int min1 = 0;
	for(int i = 1;i <= m;++i)
	{
		int g =pp[i].x;
		dp[i] = pp[i].step;//啥都沒有的是它本身一開始用了一個的
		for(int j = 1;j < i;++j)
		{
			if(pp[j].y + r <= g)//該前面有的結束時間小於這個的開始時間
			{
				dp[i] = max(dp[i],dp[j] + pp[i].step) ;//從可以遞推的前面時間遞推
			}
		}
		min1 = max(min1,dp[i]);
	}//類最長遞增子序列的思路
	printf("%d\n",min1);
	return 0;
}