POJ-3616_Milking Time

最新推荐文章于 2025-09-12 17:31:31 发布

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最新推荐文章于 2025-09-12 17:31:31 发布

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原文链接：http://www.cnblogs.com/luoxiaoyi/p/9791346.html

探讨了在给定的时间段内，通过合理安排挤奶和休息时间，Bessie如何最大化其牛奶产量的问题。使用动态规划算法解决了一个变形的上升序列问题。

Milking Time

Time Limit: 1000MS Memory Limit: 65536K

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

题意：给予M时间区间且每个区间可以挤多少牛奶，每工作一个时间段需要休息R个小时，问在N个小时最多可以得到多少牛奶。
题解：一个上升序列问题的变形。

#include <algorithm>
#include <iostream>
#include <stdio.h>
#include <cstdlib>
#include <cstring>
#include <cmath>

using namespace std;

const int maxn = 1050;

struct node
{
    int be,en,w;
}s[maxn];
int dp[maxn];

bool cmp(node a,node b)
{
    if(a.be==b.be)
        return a.en<b.en;
    return a.be < b.be;
}

int main()
{
    int n,m,i,j,r,Max,mmax;
    scanf("%d%d%d",&r,&m,&n);
    for(i=0;i<m;i++)
        scanf("%d%d%d",&s[i].be,&s[i].en,&s[i].w);
    sort(s,s+m,cmp);
    Max = 0;
    for(i=0;i<m;i++)
    {
        mmax = 0;
        for(j=0;j<i;j++)
        {
            if(s[j].en+n<=s[i].be&&mmax<dp[j])
                mmax = dp[j];
        };
        dp[i] = mmax + s[i].w;
        Max = max(Max,dp[i]);
    }
    printf("%d\n",Max);
    return 0;
}

转载于:https://www.cnblogs.com/luoxiaoyi/p/9791346.html