b

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

这个题是一个动态规划的题，线性动态规划我刚开始做的时候我感觉这个和贪心算法的题一样，因为贪心有个是活动选择，但是那个贪心选择只能选择几个项目，而不能选择出最多的价值。所以这个题要用到dp。
这个题我们对一个事物有两种选择（选或者是不选）。如果选的话那你就要查看选他的话他的前面的那个量要是什么。所以我刚开始的时候是想了一个函数和数组来记录选他的话前面能选择谁。但是这个就wa了。
然后换一个思路如果这个的开始是大于前面那个的结束那就判断前一个加上这个的价值大还是它本身的价值大，所以要在前面另dp等于这个的价值，到这里公式就出来了dp[i]=max(dp[j]+valu[i],dp[i]);
这个题有个小技巧就是对他的结束时间+r因为这个人要休息r的时间，也不用担心他会不会超过这个的截止时间因为题目中有一个是 (starting_houri < ending_houri ≤ N)

#include<iostream>
#include<algorithm>
using namespace std;
int opt[10010],dp[10010];
struct node{
    int s,d,n;
}a[1100];
int cmp(node a,node b)
{
    return a.s<b.s;
}
int main()
{
    long long n,m,r;
    cin>>n>>m>>r;
    for(int i=1;i<=m;i++)
    {
        cin>>a[i].s>>a[i].d>>a[i].n;
        a[i].d+=r;
    }
    sort(a+1,a+m+1,cmp);
    int ans=-99;
    for(int i=1;i<=m;i++)
    {
        dp[i]=a[i].n;
        for(int j=i-1;j>=1;j--)
        {
            if(a[j].d<=a[i].s)
                dp[i]=max(dp[i],dp[j]+a[i].n);
        }
        ans=max(ans,dp[i]);
    }
    cout<<ans<<endl;
}