poj3616——Milking Time（dp最大子序列）-白红宇

poj3616——Milking Time（dp最大子序列）

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发布时间：2019-05-10

本文共 1988 字，大约阅读时间需要 6 分钟。

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

给定一个时间段，在给出一些间隔，母牛会在这些间隔内产奶，求选择合适的间隔来使得产奶量最大。

#include 
   
    #include 
    
     #include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           #include 
           #include 
            
             #define INF 0x3f3f3f3f#define MAXN 1000010#define Mod 10001using namespace std;struct Node{ int x,y,sum;};Node a[MAXN];bool cmp(Node a,Node b){ if(a.x==b.x) return a.y
             
              0;--i) { dp[i]=a[i].sum; for(int j=i+1;j<=m;++j) { if(a[j].x>=a[i].y) dp[i]=max(dp[i],dp[j]+a[i].sum); } } int ans=-INF; for(int i=1;i<=m;++i) ans=max(ans,dp[i]); printf("%d\n",ans); } return 0;}

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