D

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



这题想法非常简单就是先排序，再判断第i个要不要加进时间表里（1，时间上是否符合.2，利益最大化）

#include <iostream> //这题用到滚动数组，先对时间开始进行排序，dp[i]代表前i个时间段的效率
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<algorithm>
using namespace std;
typedef long long ll;
ll dp[1100];
struct node
{
    int start,ende,eff;
}exa[1100];
bool cmp(node x,node y)
{
    if(x.start==y.start)
        return x.ende<y.ende;
    return x.start<y.start;
}
int main()
{
    int n,m,r;
    cin>>n>>m>>r;
    memset(dp,0,sizeof(dp));
    for(int i=0;i<m;i++)
    {
        cin>>exa[i].start>>exa[i].ende>>exa[i].eff;
    }
    sort(exa,exa+m,cmp);
    for(int i=0;i<m;i++) dp[i]=exa[i].eff;
    ll ans=0;
    for(int i=0;i<m;i++)//判断第i个要不要加进去
    {
        for(int j=0;j<i;j++)
        {
            if(exa[j].ende+r<=exa[i].start) dp[i]=max(dp[i],dp[j]+exa[i].eff);
        }
        ans=max(ans,dp[i]);
    }
    cout<<ans<<endl;
    return 0;
}