Machine Schedule
Time Limit: 1000MS |
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Memory Limit: 10000K |
Total Submissions: 13731 |
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Accepted: 5873 |
Description
As we all know, machine scheduling are a very classical problem in computer science and have been studied for a very long hi Story. Scheduling problems differ widely in the nature of the constraints that must be satisfied and the type of schedule desired . Here we consider a 2-machine scheduling problem.
There was machines a and B. Machine A have n kinds of working modes, which is called Mode_0, Mode_1, ..., mode_n-1, lik Ewise machine B has m kinds of working modes, MODE_0, Mode_1, ..., mode_m-1. At the beginning they is both work at Mode_0.
For k jobs given, each of the them can is processed in either one of the one of the both machines in particular mode. For example, job 0 can either is processed in machine A at mode_3 or in machine B at Mode_4, Job 1 can either be processed In machine A is mode_2 or in machine B at Mode_4, and so on. Thus, for Job I, the constraint can is represent as a triple (I, X, y), which means it can be processed either A at mode_x, or in machine B at mode_y.
Obviously, to accomplish all the jobs, we need to change the machine's working mode from time to time, but unfortunately, The machine's working mode can only is changed by restarting it manually. By changing the sequence of the jobs and assigning each job to a suitable machine, please write a program to minimize the Times of restarting machines.
Input
The input file for this program consists of several configurations. The first line of one configuration contains three positive integers:n, M (N, M <) and K (K < 1000). The following k lines give the constrains of the K jobs, each of which is a triple:i, X, Y.
The input would be terminated to a line containing a single zero.
Output
The output should is one integer per line, which means the minimal times of restarting machine.
Sample Input
5 5 100 1 11 1 22 1 33 1 44 2 15 2 26 2 37 2 48 3 39 4 30
Sample Output
3
Source
Beijing 2002 —————————————— I'm a dividing line ———————————————————————————————— a graph on water problems. The minimum overlay of the binary graph, the Hungarian algorithm. First, we construct a binary graph, the mode on each machine as the point, the a machine is left, the B machine is the right part. Connect a point between the patterns that the same task requires. After that, only the Hungarian algorithm is used to calculate the minimum coverage for DFS augmentation. Because the minimum coverage and the maximum match are equal in numerical value, only the maximum matching is required.
1#include <iostream>2#include <cstdio>3#include <cstring>4#include <cmath>5#include <algorithm>6#include <queue>7#include <cstdlib>8#include <iomanip>9 using namespacestd;Ten intRead () { One intx=0, f=1;CharCh=GetChar (); A while(ch<'0'|| Ch>'9'){if(ch=='-') f=-1; ch=GetChar ();} - while(ch>='0'&&ch<='9') {x=x*Ten+ch-'0'; ch=GetChar ();} - returnx*F; the } - intn,m,k; - intnx,ny,ans=0; - BOOLmap[101][101]; + BOOLex[101],ey[101]; - intcx[101],cy[101]; + voidChange (); A intDfsint); at intMain () - { -Std::ios::sync_with_stdio (false); - while(cin>>N) - { - if(n==0) Break; inCin>>m>>K; -memset (Map,false,sizeof(map)); to for(intI=0; i<k;i++) + { - inta,b,c; theCin>>a>>b>>C; *map[b][c]=true; $ }Panax Notoginseng Change (); -cout<<ans<<Endl; the } + return 0; A } the voidChange () + { -ans=0; $ inta,b,c; $memset (CX,0,sizeof(CX)); memset (CY,0,sizeof(CY)); - for(intI=1; i<=n;i++) - { the if(!Cx[i]) - {WuyiMemset (ex,false,sizeof(ex)); memset (EY,false,sizeof(EY)); theans+=DFS (i); - } Wu } - return; About } $ intDfsintu) - { - inta,b,c; -ex[u]=true; A intv; + for(v=1; v<=m;v++) the { - if((map[u][v]==true) && (ey[v]==false)) $ { theey[v]=true; the if(!cy[v]| |DFS (Cy[v])) the { thecx[u]=v;cy[v]=u; - return 1; in } the } the } About return 0; the}
POJ 1325
POJ 1325 Machine SCHEDULE--S.B.S.