[线性DP 子序列求和]Milking Time(POJ 3616)

[线性DP]Milking Time(POJ 3616)

​​题目链接​​

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

题意

奶牛产奶，求最多可以生产多少牛奶
给定N表示要安排接下来N小时的产奶任务.
给定M表示有M个产奶的时间段（可能有重叠的)，包括开始时间，结束时间，效益
给定R表示完成一次产奶后要休息R个小时。

思路

先对区间排序，按结束时间排。然后子序列求和DP，加个判断条件只有超过间隔R了才能状态转移
一发A了
好耶

代码

#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<math.h>
using namespace std;
int n,m,r;
typedef long long ll;
const int N=1e3+10;
struct Range
{
    int st,ed,w;
}R[N];
bool cmp(Range A,Range B)
{
    if(A.ed!=B.ed)return A.ed<B.ed;
    return 0;
}
ll f[N];
int main()
{
    scanf("%d%d%d",&n,&m,&r);
    for(int i=1;i<=m;i++)scanf("%d%d%d",&R[i].st,&R[i].ed,&R[i].w);
    sort(R+1,R+1+m,cmp);
    for(int i=1;i<=m;i++)f[i]=R[i].w;
    for(int i=1;i<=m;i++)
    {
        for(int j=0;j<i;j++)
        {
            if(R[i].st>=R[j].ed+r)f[i]=max(f[i],f[j]+R[i].w);
        }
    }
    ll ans=0;
    for(int i=1;i<=m;i++)ans=max(ans,f[i]);
    printf("%lld",ans);
    return 0;
}