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 Nhours
Sample Input
12 4 2 1 2 8 10 12 19 3 6 24 7 10 31
Sample Output
43
题意：
给出m个取奶时间段和该时间段内的取奶量，每次取奶之后需要休息r个时间段，问最大取奶量
思路：
按时间段右端点排序。
dp[i]表示第i个时间段取奶之后的最大取奶量。
#include<iostream> #include<algorithm> #include<vector> #include<stack> #include<queue> #include<map> #include<set> #include<cstdio> #include<cstring> #include<cmath> #include<ctime> #define fuck(x) cout<<#x<<" = "<<x<<endl; #define ls (t<<1) #define rs ((t<<1)+1) using namespace std; typedef long long ll; typedef unsigned long long ull; const int maxn = 100086; const int inf = 2.1e9; const ll Inf = 999999999999999999; const int mod = 1000000007; const double eps = 1e-6; const double pi = acos(-1); int n,m,r; struct node{ int l,r,v; }a[1024]; int dp[maxn]; bool cmp(node a,node b){ return a.r<b.r; } int main() { // ios::sync_with_stdio(false); // freopen("in.txt","r",stdin); scanf("%d%d%d",&n,&m,&r); for(int i=1;i<=m;i++){ scanf("%d%d%d",&a[i].l,&a[i].r,&a[i].v); } sort(a+1,a+1+m,cmp); int ans=0; for(int i=1;i<=m;i++){ dp[i]=a[i].v; for(int j=1;j<i;j++){ if(a[j].r+r<=a[i].l){dp[i]=max(dp[i],dp[j]+a[i].v);} } ans=max(ans,dp[i]); } printf("%d\n",ans); return 0; }
POJ - 3616 Milking Time (动态规划）
原文：https://www.cnblogs.com/ZGQblogs/p/10668019.html
