poj 1325 Machine Schedule 題解

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

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

——————————————————————我是華麗麗的分割線————————————————————————————————

二分圖最大點覆蓋。

最小點集覆蓋—— 用最少的點，覆蓋全部邊。

這裏面，邊指任務，點指機器模式。

這裏面的特色是在任務執行順序沒有要求的條件下才可使用最小點集覆蓋。

這裏具體仍是使用匈牙利算法，找到最大匹配的邊數——實際意義爲找到匹配邊的其中三個對應點，可以覆蓋全部的任務。

 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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