poj 3616(动态规划)

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Milking Time Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 7265 Accepted: 3043 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 ≤…

POJ - 3616 Milking Time (动态规划）

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 ha…

动态规划：POJ 3616 Milking Time

#include <iostream> #include <algorithm> #include <cstring> #include <cstdio> #include <cstdlib> using namespace std; int N, M, R; // N个小时, M个时间间隔, R个休息时间 + ; struct Cow { int start; int end; int value; Cow(, , ) : start(s),…

【POJ - 3616】Milking Time（动态规划）

Milking Time 直接翻译了 Descriptions 贝茜是一个勤劳的牛.事实上,她如此专注于最大化她的生产力,于是她决定安排下一个N(1≤N≤1,000,000)小时(方便地标记为0..N-1),以便她生产尽可能多的牛奶. 农民约翰有一个M(1≤M≤1,000)可能重叠的间隔列表,他可以在那里进行挤奶.每个区间我有一个起始小时(0≤starting_houri≤N),一个结束小时(starting_houri <ending_houri≤N),以及相应的效率(1≤efficien…

POJ 3616 Milking Time（加掩饰的LIS）

传送门: http://poj.org/problem?id=3616 Milking Time Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 13406 Accepted: 5655 Description Bessie is such a hard-working cow. In fact, she is so focused on maximizing her productivity that she dec…

POJ 3616 Milking Time (排序+dp)

题目链接:http://poj.org/problem?id=3616 有头牛产奶n小时(n<=1000000),但必须在m个时间段内取奶,给定每个时间段的起始时间和结束时间以及取奶质量且两次取奶之间须间隔r-1个小时,求最大取奶质量也就是说r = 2时 3分结束取奶,至少在5分才能取. 按照时间排序,dp[i]表示i时段的最大产奶量 //#pragma comment(linker, "/STACK:102400000, 102400000") #include <a…

POJ 3616 Milking Time（最大递增子序列变形）

题目链接:http://poj.org/problem?id=3616 题目大意:给你时间N,还有M个区间每个区间a[i]都有开始时间.结束时间.生产效率(时间都不超过N),只能在给出的时间段内生产,要求合理安排时间求出最大生产价值. 解题思路:把区间按开始时间排序,于是有状态转移方程:dp[i]=max(dp[i],dp[j]+a[i].val)(前提是a[j].end+r<=a[i].start,i是区间的序号,j是i前面的区间) 相当于最大递增子序列的变形,写法差不多. 代码: #incl…