題目：

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_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

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

題意：有m個時間段，每段時間都有開始時間，結束時間，產生價值三個屬性。John可以選擇一些時間段來產生價值，當然選擇的時間段時間不能有重複，並且進行完一個時間段的工作之後，他必須休息r時間。求從這m個時間段中產生價值的最大值。

思路：首先要將所有時間段按照開始時間進行從小到大的排序。

之後定義dp[i]為：假設第i個時間段進行產奶時時，產生價值的總和的最大值。

狀態轉移方程為： dp[i] = max(dp[i] , dp[i]+list[j].value) ；( j<i , list[j].end+r<=list[i].start)

也就是說如果當前時間時間段產奶，產生價值的最大值為當前時間段產生的價值加上之前時間段所能產生的最大值。

而結果自然是所有假設中的最大值。

程式碼：

#include<cstdio>  
#include<cstdlib>  
#include<cmath>  
#include<cstring>  
#include<iostream>  
#include<algorithm>  
#include<queue>  
#include<map>  
#include<stack> 
#include<set>
#define sd(x) scanf("%d",&x)
#define ss(x) scanf("%s",x)
#define sc(x) scanf("%c",&x)
#define sf(x) scanf("%f",&x)
#define slf(x) scanf("%lf",&x)
#define slld(x) scanf("%lld",&x)
#define me(x,b) memset(x,b,sizeof(x))
#define pd(d) printf("%d\n",d);
#define plld(d) printf("%lld\n",d);
#define eps 1.0E-8
// #define Reast1nPeace

typedef long long ll;

using namespace std;

const int INF = 0x3f3f3f3f;

struct fj{
	int s,e;
	int v;
}lis[1010];

bool cmp(fj x , fj y){
	if(x.s == y.s){
		return x.e < y.e;
	}
	return x.s < y.s;
}

int dp[1010];

int main(){
#ifdef Reast1nPeace
	freopen("in.txt", "r", stdin);
	freopen("out.txt", "w", stdout);
#endif
	ios::sync_with_stdio(false);
	int n,m,r;
	cin>>n>>m>>r;
	for(int i = 0 ; i<m ; i++){
		cin>> lis[i].s >> lis[i].e >> lis[i].v;
	}
	sort(lis , lis+m ,  cmp);
	
	int maxn = 0;
	for(int i = 0 ; i<m ; i++){
		dp[i] = lis[i].v;
		for(int j = 0 ; j<i ; j++){
			if(lis[j].e+r <= lis[i].s){
				dp[i] = max(dp[i] , dp[j]+lis[i].v);
			}
		}
		maxn = max(dp[i] , maxn);
	}
	cout<<maxn<<endl;
	return 0;
}