POJ3616 Milking Time

POJ3616 Milking Time

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_houriN), an ending hour (starting_houri < ending_houriN), 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 ≤ RN) 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

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2
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12 4 2
1 2 8
10 12 19
3 6 24
7 10 31

Sample Output

1
43

Answer

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#include <cstdio>
#include <algorithm>
using namespace std;

struct Cow{
int s, t, k;
Cow() {}
Cow(int s, int t, int k) : s(s), t(t), k(k) {}
bool operator<(const Cow& c) const {
return t < c.t;
}
} cows[1010];

int main() {
int n, m, r, dp[1010];
scanf("%d %d %d", &n, &m, &r);
for (int i = 0; i < m; ++i) {
int st, et, k;
scanf("%d %d %d", &st, &et, &k);
cows[i] = Cow(st, et+r, k);
}

sort(cows, cows+m);
for (int i = 0; i < m; ++i) {
dp[i] = cows[i].k;
for (int j = 0; j < i; ++j)
if (cows[j].t <= cows[i].s)
dp[i] = max(dp[i], dp[j] + cows[i].k);
}

int ans = 0;
for (int i = 0; i < m; ++i)
ans = max(ans, dp[i]);
printf("%d\n", ans);
return 0;
}

Analysis

  1. 若干个时间区间,每个区间一个价值,且选中的区间之间至少间隔 r,求最大价值。
  2. 将区间按照终点大小排序,以第 i 个区间为结尾的最大值 dp[i] = max {dp[j], j <i,且 j 满足条件} + a[i].value。
  3. 最后查看 dp 中的最大值。
# POJ

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