动态规划 POJ3616 Milking Time

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
struct cow
{
    int start;
    int endd;
    int price;
};
bool cmp(cow a,cow b)
{
    return a.start<b.start;
}
int main()
{
    int n,m,r;
    cow a[1005];
    cin>>n>>m>>r;
    int i,j;
    for(i=0;i<m;i++)
    {
        cin>>a[i].start>>a[i].endd>>a[i].price;
        a[i].endd+=r;
    }
    sort(a,a+m,cmp);
    int dp[1005]={0};
    for(i=0;i<m;i++)
    {
        dp[i]=a[i].price;
        for( j=0;j<i;j++)
        {
            if(a[i].start>=a[j].endd)
            {
                dp[i]=max(dp[i],dp[j]+a[i].price);
            }
        }
    }
    cout<<*max_element(dp,dp+m)<<endl;
    return 0;
}

　　

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

思路很简单，但感觉不是很好想，一开始想成背包问题了= =!

言归正传，这题就先按开始时间排序，按最长上升序列O（n^2）的方法解，状态转移方程为dp[i]=max(dp[i],dp[j]+a[i].price)，0<=j<i&&a[i].start>=a[j].end.