POJ 1325 Machine Schedule——S.B.S.

Machine Schedule

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 13731		Accepted: 5873

Description

As we all know, machine scheduling is a very classical problem in computer science and has been studied for a very long history. 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 are two machines A and B. Machine A has n kinds of working modes, which is called mode_0, mode_1, ..., mode_n-1, likewise machine B has m kinds of working modes, mode_0, mode_1, ... , mode_m-1. At the beginning they are both work at mode_0. 

For k jobs given, each of them can be processed in either one of the two machines in particular mode. For example, job 0 can either be processed in machine A at mode_3 or in machine B at mode_4, job 1 can either be processed in machine A at mode_2 or in machine B at mode_4, and so on. Thus, for job i, the constraint can be represent as a triple (i, x, y), which means it can be processed either in machine 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 be 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 < 100) and k (k < 1000). The following k lines give the constrains of the k jobs, each line is a triple: i, x, y. 

The input will be terminated by a line containing a single zero. 

Output

The output should be one integer per line, which means the minimal times of restarting machine.

Sample Input

5 5 10
0 1 1
1 1 2
2 1 3
3 1 4
4 2 1
5 2 2
6 2 3
7 2 4
8 3 3
9 4 3
0

Sample Output

3

Source

Beijing 2002

——————————————我是分割线————————————————————————————————

一道图论水题。

二分图最小覆盖，匈牙利算法。

首先构造二分图，各机器上的模式作为点，A机器为左部，B机器为右部。

把同一任务需要的模式之间连点。

之后只需用匈牙利算法计算，dfs增广求最小覆盖。

又因为二分图最小覆盖与最大匹配在数值上相等，只需求最大匹配即可。

 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 namespace std;
10 int read(){
11     int x=0,f=1;char ch=getchar();
12     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
13     while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
14     return x*f;
15 }
16 int n,m,k;
17 int nx,ny,ans=0;
18 bool map[101][101];
19 bool ex[101],ey[101];
20 int cx[101],cy[101];
21 void change();
22 int dfs(int);
23 int main()
24 {
25     std::ios::sync_with_stdio(false);
26     while(cin>>n)
27     {
28         if(n==0) break;
29         cin>>m>>k;
30         memset(map,false,sizeof(map));
31         for(int i=0;i<k;i++)
32         {
33             int a,b,c;
34             cin>>a>>b>>c;
35             map[b][c]=true;
36         }
37         change();
38         cout<<ans<<endl;
39     }
40     return 0;
41 }
42 void change()
43 {
44     ans=0;
45     int a,b,c;
46     memset(cx,0,sizeof(cx));memset(cy,0,sizeof(cy));
47     for(int i=1;i<=n;i++)
48     {
49         if(!cx[i])
50         {
51             memset(ex,false,sizeof(ex));memset(ey,false,sizeof(ey));
52             ans+=dfs(i);
53         }
54     }
55     return;
56 }
57 int dfs(int u)
58 {
59     int a,b,c;
60     ex[u]=true;
61     int v;
62     for(v=1;v<=m;v++)
63     {
64         if((map[u][v]==true)&&(ey[v]==false))
65         {
66             ey[v]=true;
67             if(!cy[v]||dfs(cy[v]))
68             {
69                 cx[u]=v;cy[v]=u;
70                 return 1;
71             }
72         }
73     }
74     return 0;
75 }

POJ 1325