ZOJ3554 A miser Boss (DP)

Reprint Please specify source: http://www.cnblogs.com/fraud/--by fraud

A Miser Boss Time limit: 2 Seconds Memory Limit: 65536 KB

There is three different lathes in a factory namely A,b,c. They is all able to work on all kinds of workpieces. And there is N workpieces to being processed, denoted by ,.., n.

Due to different internal implemetations, for a specific workpiece I, it costs a a[i] seconds to finish The job on it while B takes B[i] and C takes C[i] seconds. Each lathe can work on at the most one workpiece a time.

Additionally, the Boss is so stingy that he didn ' t want his lathes to being disengaged at any time before the end of the work . That is:
1. There should not being a time when one of the lathes are in leisure.
2. All lathes should stop Simultaneously,and start from time 0.

Your task is to find if there exists a arrangement meeting with the constraints above. If there is,output the minimal time needed to process all the workpieces. Output a line NO else.

Input

There is multiple test cases. The first line of all case are an integer N (0 < n≤40) , followed by N lines. In the following N lines, line i contains three integer,A[i],b[i],c[i] (0 < a[i],b[i],c[i]≤120 && 0 < sum (a[i]), SUM (b[i]), SUM (C[i]) ≤120) indicating the seconds a,b,c takes to process workpiece I .

Output

Output one line the minimal seconds it takes to process all workpieces within the constraints if there are an arrangement. Print NO if not.

Sample Input

37 1 21 7 11 3 721 2 33 2 1

Sample Output

1NO

Forgive me stupid, can only think of n*120*120*120 method, Dp[i][j][k][l] said to take the first I when a take the sum is J,b is K,c is L this scheme is feasible

As long as the judgment is feasible, then in fact, the four-dimensional can be optimized. So I think I can use bitset, every time just or a little, yes, it's so simple and rude.

1 /**2 * Code generated by Jhelper3 * More info:Https://github.com/AlexeyDmitriev/JHelper4 * @author Xyiyy @Https://github.com/xyiyy5  */6 7#include <iostream>8#include <fstream>9 Ten //##################### One //Author:fraud A //Blog:http://www.cnblogs.com/fraud/ - //##################### - //#pragma COMMENT (linker, "/stack:102400000,102400000") the#include <iostream> -#include <sstream> -#include <ios> -#include <iomanip> +#include <functional> -#include <algorithm> +#include <vector> A#include <string> at#include <list> -#include <queue> -#include <deque> -#include <stack> -#include <Set> -#include <map> in#include <cstdio> -#include <cstdlib> to#include <cmath> +#include <cstring> -#include <climits> the#include <cctype> *  $ using namespacestd;Panax Notoginseng #defineRep (X, N) for (int x=0; x<n; X + +) - #defineREP2 (X, L, R) for (int x=l; x<=r; X + +) the  + inta[ the], b[ the], c[ the]; A  the#include <bitset> +  -bitset<121> dp[ the][121][121]; $  $ classTASKF { -  Public: -     voidSolve (Std::istream &inch, Std::ostream & out) { the         intN; -          while(inch>>N) {Wuyi             inttot =0; theREP2 (I,1, N) { -                 inch>> A[i] >> B[i] >>C[i]; WuTot + =C[i]; -             } AboutRep (i, n +1) $Rep (J,121) -Rep (K,121) Dp[i][j][k].reset (); -             intTA =0, TB =0; -dp[0][0][0][0] =1; AREP2 (I,1, N) { +Rep2 (J,0, TA) { theRep2 (k,0, TB) { -Dp[i][j + a[i]][k] |= dp[i-1][j][k]; $Dp[i][j][k + b[i]] |= dp[i-1][j][k]; theDp[i][j][k] |= (Dp[i-1][J][K] <<c[i]); the                     } the                 } theTa + =A[i]; -TB + =B[i]; in             } the             intAns =0; theREP2 (I,1, -) { About                 if(Dp[n][i][i][i]) { theAns =i; the                      Break; the                 } +             } -             if(ANS) out<< ans <<Endl; the             Else  out<<"NO"<<Endl;Bayi         } the     } the }; -  - intMain () { theStd::ios::sync_with_stdio (false); theStd::cin.tie (0); the TASKF solver; theStd::istream &inch(std::cin); -Std::ostream & out(std::cout); theSolver.solve (inch, out); the     return 0; the}

ZOJ3554 A miser Boss (DP)