【动态规划】POJ-3616

一、题目

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

二、思路&心得

• 对于”每头奶牛挤奶后需要休息R分钟“的条件限制，直接让每头奶牛的结束时间加上R分钟，来消除这个限制。
• 定义挤奶的结构体（开始时间、结束时间、获得收益），然后按照结束时间从小到大进行排序。
• 这题有点像贪心中的区间问题，排完序后,定义dp[i]为：，到dp[i].end这个时间点为止，可以获得的最大收益。显然，dp[i]满足如下递推式：dp[i] = dp[k] + T[i].gollon。（其中k为dp[0 to i - 1]的最大值下标，且T[k].end <= T[i].start）
• 注意题目不不是在最后一个时间点取到最大值，因此需要记录最大的dp。

三、代码

#include<cstdio>
#include<algorithm>
using namespace std;
const int MAX_M = 1005;

int N, M, R;

int dp[MAX_M];

struct times {
	int start;
	int end;
	int gollon;
} T[MAX_M];

bool cmp(times a, times b) {
	return a.end < b.end;
}

int main() {
	int maxGollons = 0;
	scanf("%d %d %d", &N, &M, &R);
	for (int i = 0; i <= M; i ++) {
		scanf("%d %d %d", &T[i].start, &T[i].end, &T[i].gollon);
		T[i].end += R;
	}
	sort(T, T + M, cmp);
	for (int i = 0; i < M; i ++) {
		dp[i] = T[i].gollon;
		for (int j = 0; j < i; j ++) {
			if (T[j].end <= T[i].start) {
				dp[i] = max(dp[i], dp[j] + T[i].gollon);
			}
		}
		maxGollons = max(maxGollons, dp[i]);
	}
	printf("%d\n", maxGollons);
	return 0;
}