"Bzoj 1642" [Usaco2007 nov]milking time Milking times

1642: [Usaco2007 nov]milking time Milking Hour: 5 Sec Memory Limit:
Submit: 590 Solved: 337
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Bessie was a very hard-working cow, and she always focused on improving her own production. In order to produce more milk, she expected the next N (1≤n≤1,000,000) hours, labeled 0. N-1. Farmer John planned to 1≤m≤1,000 a time period for milking. Each time period has a start time (0≤ start time ≤n), and an end time (start time < end time ≤n), and a yield (1≤ yield ≤1,000,000) represents the amount of milk that can be milked from Bessie. Farmer John is milking from the start time to the end time respectively. Each milking must use the entire time period. But even Bessie had her production limits. After each milking, she must rest R (1≤r≤n) hours before milking next time. Given the time period of the farmer John Program, please calculate the maximum amount of milking in N hours.

Input

Section1row of three integersN,M,R.NextMrows, three integers per lineSi,Ei,Pi.

Output

the maximum amount of milk produced.

Sample Input12 4 2
1 2 8
10 12 19
3 6 24
7 10 31
Sample Output +HINT

Note: End time not milking

Dp.

Sort by start and end time, then DP:

F[i] Indicates the maximum amount of milk to be extruded using the first I milking time.

F[i]=max (F[k]+w[i]) (A[K].E+R<=A[I].S)

Notice the end time not milking!! means this is <=.

#include <iostream> #include <cstdio> #include <algorithm> #include <cstring> #include < Cstdio>using namespace Std;int n,m,r;struct data{int s,e,v;} A[1005];int f[1005];bool cmp (data a,data b) {if (A.S==B.S) return A.e<b.e;return A.S<B.S;} int main () {    scanf ("%d%d%d", &n,&m,&r), for (int i=1;i<=m;i++) scanf ("%d%d%d", &a[i].s,&a[i ].E,&A[I].V); sort (a+1,a+1+m,cmp); f[1]=a[1].v;int ans=f[1];for (int i=2;i<=m;i++) {f[i]=a[i].v;for (int j=1;j <i;j++) if (A[J].E+R<=A[I].S) F[i]=max (F[I],F[J]+A[I].V); Ans=max (F[i],ans);} Cout<<ans<<endl;return 0;}

"Bzoj 1642" [Usaco2007 nov]milking time Milking times