POJ 3616Milking Time

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

题解：
　　这个题目还是比较水的吧，设dp[i]表示选以i结尾的物品的最大价值。
　　那么dp[i]=max(dp[j]+v[i])(l[i]-r[j]>=休息时间)。

代码：

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <iostream>
#define MAXN 2000
#define ll long long
using namespace std;
struct qvjian{
    int l,r,v;
    void read(){
        scanf("%d%d%d",&l,&r,&v);
    }
}a[MAXN*2];
ll dp[MAXN];
int h,n,k;

bool cmp(qvjian x,qvjian y){
    return x.r<y.r;
}

int main()
{
    scanf("%d%d%d",&h,&n,&k);
    for(int i=1;i<=n;i++) a[i].read();
    sort(a+1,a+n+1,cmp);
    memset(dp,0,sizeof(dp));
    a[0].r=-(1<<30);
    for(int i=1;i<=n;i++){
        for(int j=i-1;j>=0;j--){
            if(a[i].l-a[j].r>=k)
                dp[i]=max(dp[i],dp[j]+a[i].v);
        }
    }
    ll ans=0;
    for(int i=1;i<=n;i++) ans=max(ans,dp[i]);
    printf("%lld
",ans);
    return 0;
}