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RMilkingTimePOJ3616(DP)

R-MilkingTimeTimeLimit:1000MSMemoryLimit:65536KB64bitIOFormat:%I64d&%I64uSubmitStatusP

R - Milking Time
Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
Submit Status Practice POJ 3616

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_houri ≤ N), an ending hour (starting_houri
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

先把读入的区间从小到大排序,然后dp



#include
#include
#include
using namespace std;
const int maxn=10005;
struct node
{
    int l,r,data;
    bool operator<(const node& tmp) const{
        if(l!=tmp.l)
            return l=r){
                maxx=max(maxx,dp[j]);//找到这个区间前面的最大值
            }
            dp[i]=maxx+p[i].data;
        }
       printf("%d\n",*max_element(dp,dp+m+1));
    }
    return 0;
}















R - Milking Time POJ 3616 ( DP )


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