POJ1325 Machine Schedule

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

 

 

正解：二分图匹配

解题报告：

　　今天考试T3，今天也就切了这道题了...

　　考虑每个点相当于是一个矩阵里面的坐标，要么这一行被选，要么这一列被选，所以可以直接转成最小覆盖集模型。

　　首先，把坐标含0的全部去掉，然后跑二分图最大匹配，由于最小覆盖等于最大匹配，可以直接得到答案。

　　所以这道题就是一个最小点覆盖的裸题...

 

 

 //It is made by jump~
 #include <iostream>
 #include <cstdlib>
 #include <cstring>
 #include <cstdio>
 #include <cmath>
 #include <algorithm>
 #include <ctime>
 #include <vector>
 #include <queue>
 #include <map>
 #include <set>
 using namespace std;
 typedef long long LL;
 const int MAXN = ;
 const int MAXM = ;
 int n,m,k,ecnt;
 int first[MAXN],to[MAXM],next[MAXM],match[MAXN];
 bool vis[MAXN];
 int ans;

 inline int getint()
 {
        int w=,q=; char c=getchar();
        while((c<'' || c>'') && c!='-') c=getchar(); if(c=='-') q=,c=getchar();
        while (c>='' && c<='') w=w*+c-'', c=getchar(); return q ? -w : w;
 }

 inline bool dfs(int x){
     if(vis[x]) return false;
     vis[x]=;
     for(int i=first[x];i;i=next[i]) {
     int v=to[i]; if(vis[v])  continue;
     if(!match[v] || dfs(match[v])) {
         match[x]=v; match[v]=x;
         return true;
     }
     }
     return false;
 }

 inline void work(){
     while(scanf("%d",&n)!=EOF) {
     if(n==) break;
     m=getint(); k=getint();
     int x,y; memset(first,,sizeof(first)); memset(match,,sizeof(match));
     ecnt=;
     for(int i=;i<=k;i++) {
         x=getint(); x=getint(); y=getint();
         if(x== || y==) { i--; k--; continue; }
         y+=n;
         next[++ecnt]=first[x]; first[x]=ecnt; to[ecnt]=y;
         next[++ecnt]=first[y]; first[y]=ecnt; to[ecnt]=x;
     }
     ans=;
     for(int i=;i<n;i++) {
         memset(vis,,sizeof(vis));
         if(dfs(i)) ans++;
     }
     printf("%d\n",ans);
     }
 }

 int main()
 {
   work();
   return ;
 }