【POJ】3616 Milking Time（dp）

Milking Time

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10898		Accepted: 4591

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

Source

USACO 2007 November Silver

 

 

 

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分析：这题看到题首先想到的是O(nm)做法，即用dp[i][j]表示i到j的最大产奶数。然而这种做法针对这道题目，连数组都开不下，更别提时间了。

　　正确做法是这样子的，因为m比较小，所以先按照时间进行排序，然后可以用dp[i]表示前i次产奶的最大值，

　　递推关系：第i个时间段挤奶的最大值 = 前 i – 1 个时间段挤奶最大值中的最大值 + 第i次产奶量。

 

 

#include <cstdio>
#include <algorithm>
using namespace std;
int dp[1000007];
struct c{
    int begin,end,e;
}a[1000007];
bool cmp(c a,c b)
{
    if(a.begin<b.begin) return true;
    if(a.begin==b.begin&&a.end<b.end) return true;
    return false;
}
int main()
{
    int n,m,r;
    scanf("%d%d%d",&n,&m,&r);
    for(int i=1;i<=m;i++)
    {
        scanf("%d%d%d",&a[i].begin,&a[i].end,&a[i].e);
        a[i].end+=r;// 实际的结束时间还需要加上休息时间
    }
    sort(a+1,a+1+m,cmp);
    for(int i=1;i<=m;i++)
    {
        dp[i]=a[i].e;
        for(int j=1;j<=i;j++)
        {
            if(a[j].end<=a[i].begin)
            dp[i]=max(dp[i],dp[j]+a[i].e);
        }
    }
    printf("%d",*max_element(dp,dp+1+m));
    return 0;
}