Milking Time

 

Milking Time

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

Submit   Status   Practice   POJ 3616

Appoint description:   System Crawler  (2015-04-30)

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_hour_i ≤ N), an ending hour (starting_hour_i <<SPAN class=Apple-converted-space> 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

一开始想了三维的方程，实在是太复杂，而且不可能实现，最后发现只需要一维的状态转移，对于时间段的动规还是不熟，其实还是取与不取的状态转移。

 

 

if(st[j]>=ed[i]+r)

dp[j]=max(dp[j],dp[i]+eff[j]);

 

#include
#include
#include
#include
using namespace std;
int dp[1005];
struct Cow{
 int st;
 int ed;
 int eff;
}cow[1005];
int max(int a, int b)
{
 return a > b ? a : b;
}
bool cmp(const Cow&a, const Cow&b)
{
 if (a.ed == b.ed)return a.st < b.st;
 return a.ed < b.ed;
}
int main()
{
 ios::sync_with_stdio(false);
 int n, m, r, i, j, ans;
 cin >> n >> m >> r;
 for (i = 1; i <= m; i++)
  cin >> cow[i].st >> cow[i].ed >> cow[i].eff;
 sort(cow + 1, cow + 1 + m, cmp);
 ans = 0;
 for (i = 1; i <= m; i++)
  dp[i] = cow[i].eff;
 for (i = 1; i <= m; i++)
 {
  for (j = i + 1; j <= m; j++)
  {
   if (cow[j].st >= cow[i].ed + r)
    dp[j] = max(dp[j], dp[i] + cow[j].eff);
  }
  ans = max(ans, dp[i]);
 }

 cout << ans << "\n";
 return 0;
}