POJ3616：Milking Time

Milking Time

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5682		Accepted: 2372

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 <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

题意：在各段时间之内产牛奶的数量不同，有休息时间，过了休息时间之后才能继续产牛奶。问总共能产多少牛奶。

代码：

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

struct node{
	int start;
	int end;
	int ef;
}eff[1005];

int dp[1000005];
bool cmp(struct node node1,struct node node2)
{
	if(node1.start==node2.start)
		return node1.end<node2.end;
	else
		return node1.start<node2.start;
}

int main()
{
	int N,M,R;
	cin>>N>>M>>R;

	int i,j;
	for(i=1;i<=M;i++)
	{
		cin>>eff[i].start>>eff[i].end>>eff[i].ef;
		eff[i].end+=R;
	}
	sort(eff+1,eff+M+1,cmp);

	memset(dp,0,sizeof(dp));

	for(i=1;i<=M;i++)
	{
		dp[eff[i].end]=eff[i].ef;
	}

	int max_i;
	for(i=1;i<=M;i++)
	{
		max_i=0;
		for(j=1;j<i;j++)
		{
			if(eff[j].end<=eff[i].start)
			{
				if(dp[eff[j].end]>max_i)
				{
					max_i=dp[eff[j].end];
				}
			}
		}
		dp[eff[i].end]=max(max_i+eff[i].ef,dp[eff[i].end]);
	}
	max_i=0;
	for(i=1;i<=M;i++)
	{
		if(dp[eff[i].end]>max_i)
		{
			max_i=dp[eff[i].end];
		}
	}
	cout<<max_i<<endl;

	return 0;
}

版权声明：本文为博主原创文章，未经博主允许不得转载。