POJ 3616 D - Milking Time(DP)

最新推荐文章于 2021-08-30 14:52:22 发布

原创最新推荐文章于 2021-08-30 14:52:22 发布 · 492 阅读

1 ·

CC 4.0 BY-SA版权

DP 专栏收录该内容

24 篇文章

订阅专栏

本文介绍了一种算法解决方案，帮助一头名为Bessie的奶牛通过合理安排挤奶时间和休息时间，在给定的时间段内实现最大的牛奶产量。算法通过动态规划方法，考虑了挤奶效率和必要的休息周期。

Milking Time Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%lld & %llu

Submit Status

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

Sample Output

首先按开始时间排序，然后

f[i]表示前i个最多能够挤得的奶，那么有

f[i]=w[i]，初始化

f[i]=max(f[i], f[j]+w[i]) 当j的结束时间+R不超过i的开始时间

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int dp[1010];
int n,m,r;
struct node
{
    int s,e,v;
}a[1010];
bool cmp(node a,node b){
    return a.s<b.s ||(a.s==b.s&&a.e<b.e);
}
int main()
{
    while(~scanf("%d %d %d",&n,&m,&r))
    {
        for(int i=0;i<m;i++)
            scanf("%d %d %d",&a[i].s,&a[i].e,&a[i].v);
        sort(a,a+m,cmp);
        memset(dp,0,sizeof(dp));
        for(int i=0;i<m;i++)
            dp[i]=a[i].v;
        int ans=0;
        for(int i=0;i<m;i++){
            for(int j=0;j<i;j++)
            {
                if(a[j].e+r<=a[i].s)
                    dp[i]=max(dp[i],dp[j]+a[i].v);
            }
            ans=max(dp[i],ans);
        }
        printf("%d\n",ans);
    }
    return 0;
}