POJ 3616 Milking time (sort +DP)

Title Link: http://poj.org/problem?id=3616

Cows produce n hours (n<=1000000), but milk must be taken within m time periods, given the start and end times of each period and the quality of the milk taken.

And two times between the milk to be spaced r-1 hours, to maximize the quality of milk intake

That is, R = 2:3 to the end of the milk, at least 5 minutes to take.

In chronological order, Dp[i] represents the maximum amount of milk produced during the I period

1 //#pragma COMMENT (linker, "/stack:102400000, 102400000")2#include <algorithm>3#include <iostream>4#include <cstdlib>5#include <cstring>6#include <cstdio>7#include <vector>8#include <cmath>9#include <ctime>Ten#include <list> One#include <Set> A#include <map> - using namespacestd; -typedefLong LongLL; thetypedef pair <int,int>P; - Const intN = 1e3 +5; - structData { -     intL, R, Val; +     BOOL operator< (Constdata& CMP)Const { -         returnL <CMP.L; +     } A }a[n]; at intDp[n]; -  - intMain () - { -     intN, M, R; -      while(~SCANF (" %d%d%d", &n, &m, &r)) { inMemset (DP,0,sizeof(DP)); -          for(inti =1; I <= m; ++i) { toscanf" %d%d%d", &AMP;A[I].L, &AMP;A[I].R, &a[i].val); +         } -Sort (A +1, A + M +1); the         intres =0; *          for(inti =1; I <= m; ++i) { $Dp[i] =A[i].val;Panax Notoginseng              for(intj =1; J < I; ++j) { -                 if(A[I].L-A[J].R >=r) { theDp[i] = max (Dp[i], dp[j] +a[i].val); +                 } A             } theres =Max (res, dp[i]); +         } -printf"%d\n", res); $     } $     return 0; -}

POJ 3616 Milking time (sort +DP)