poj 1325 Machine Schedule 题解

Machine Schedule

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 14479		Accepted: 6172

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

Sample Output

Source

Beijing 2002

——————————————————————我是华丽丽的分割线————————————————————————————————

二分图最大点覆盖。

最小点集覆盖—— 用最少的点，覆盖所有边。

这里面，边指任务，点指机器模式。

这里面的特点是在任务执行顺序没有要求的条件下才可以使用最小点集覆盖。

这里具体还是使用匈牙利算法，找到最大匹配的边数——实际意义为找到匹配边的其中三个对应点，能够覆盖所有的任务。

 1 /*
 2     Problem:
 3     OJ:     
 4     User:   
 5     Time:   
 6     Memory:  
 7     Length:  
 8 */
 9 #include<iostream>
10 #include<cstdio>
11 #include<cstring>
12 #include<cmath>
13 #include<algorithm>
14 #include<queue>
15 #include<cstdlib>
16 #include<iomanip>
17 #include<cassert>
18 #include<climits>
19 #include<vector>
20 #include<list>
21 #include<map>
22 #define maxn 101
23 #define F(i,j,k) for(int i=j;i<=k;i++)
24 #define M(a,b) memset(a,b,sizeof(a))
25 #define FF(i,j,k) for(int i=j;i>=k;i--)
26 #define inf 0x7fffffff
27 #define maxm 2016
28 #define mod 1000000007
29 //#define LOCAL
30 using namespace std;
31 int read(){
32     int x=0,f=1;char ch=getchar();
33     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
34     while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
35     return x*f;
36 }
37 int n,m,k;
38 int mp[maxn][maxn];
39 int px[maxn],py[maxn];
40 int ans;
41 int cx[maxn],cy[maxn];
42 inline int path(int u)
43 {
44     cx[u]=1;
45     F(i,1,m){
46         if(mp[u][i]>0&&!cy[i]){
47             cy[i]=1;
48             if(!py[i]||path(py[i]))
49             {
50                 px[u]=i;
51                 py[i]=u;
52                 return 1;
53             }
54         }
55     }
56     return 0;
57 }
58 inline void solve()
59 {
60     ans=0;
61     M(px,0);M(py,0);
62     F(i,1,n){
63         if(!px[i]){
64             M(cx,0);M(cy,0);
65             ans+=path(i);
66         }
67     }
68     return;
69 }
70 int main()
71 {
72     std::ios::sync_with_stdio(false);//cout<<setiosflags(ios::fixed)<<setprecision(1)<<y;
73     #ifdef LOCAL
74     freopen("data.in","r",stdin);
75     freopen("data.out","w",stdout);
76     #endif
77     while(cin>>n)
78     {
79         if(n==0) break;
80         cin>>m>>k;M(mp,0);
81         F(i,1,k){
82             int a,b,c;
83             cin>>a>>b>>c;
84             mp[b][c]=1;
85         }
86         solve();
87         cout<<ans<<endl;
88     }
89     return 0;
90 }

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