POJ3616 Milking Time DP

Milking Time

Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

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 < 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

문제 풀이:

먼저 젖소의 시작 근무 시간에 따라 순서를 정하고 dp[i]는 0에서 i까지의 최대 수익을 나타낸다. 그러면 dp[i]=max(dp[i], dp[j]+cow[i].effic);

코드

#include <iostream>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<cstdio>
using namespace std;

int dp[1000005];
struct milk
{
   int start_hour;
   int end_hour;
   int effic;
}cow[1001];

bool cmp(milk a,milk b)
{
   return a.start_hour < b.start_hour;
}
int main()
{
    int n,m,r;
    scanf("%d%d%d",&n,&m,&r);
    for (int i = 0;i<m;i++)
      {
        scanf("%d%d%d",&cow[i].start_hour,&cow[i].end_hour,&cow[i].effic);
        cow[i].end_hour+=r;                  //      
      }
     sort(cow,cow+m,cmp);
     for (int i = 0;i<m;i++)
     {
        dp[i] = cow[i].effic;
        for (int j = 0;j<i;j++)
          if (cow[i].start_hour >= cow[j].end_hour)
         dp[i] = max(dp[i],dp[j]+cow[i].effic);
     }

     int maxn = 0;
     for (int i = 0;i<m;i++)
        if (dp[i] > maxn)
          maxn = dp[i];
     printf("%d
",maxn);
    return 0;
}