Machine Schedule（最小点覆盖=最大匹配）
Link：http://poj.org/problem?id=1325
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 12012 | Accepted: 5118 |
Description
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 will be terminated by a line containing a single zero.
Output
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
分析部分内容来自http://blog.csdn.net/hackbuteer1/article/details/7398008
分析：显然，机器重启次数是两台机器需要使用的不同模式的个数。把每个任务看成一条边，即A机器的每个模式看成一个X节点，B机器的每个模式看成一个Y节点，任务i为边（ai, bi）。本题即求最少的点让每条边至少与其中的一点关联，即求一个点的最小覆盖。可以证明，这个最小覆盖就是该二分图的最大匹配数。故二分图匹配的模型就建好了。注意到开始时机器都处于0模式，所以如果某个任务可以在0模式下执行，则我们可以不考虑该任务，假定它已经被完成即可，也就是建图的时候不要把与0关联的边加到二分图中就可以得到正确的解。
思想：A，B两台机器的所有模式（除0模式）分别成为二分图的二边，能完成同一份工作的有边连接。如果一个工作有一台的0模式能完成，不需要加到二分图中。用匈牙利算法求出最大匹配，由最小点覆盖=最大匹配。
#include<iostream>
#include<algorithm>
#include<vector>
#include<cstring>
using namespace std;
const int maxn=105;
vector<int>map[maxn];
bool vis[maxn];
int match[maxn],n,m,k,ans;
bool dfs(int u)
{
for(int i=0;i<map[u].size();i++)
{
if(!vis[map[u][i]])
{
vis[map[u][i]]=true;
if(match[map[u][i]]==-1||dfs(match[map[u][i]]))
{
match[map[u][i]]=u;
return true;
}
}
}
return false;
}
void hangry()
{
memset(match,-1,sizeof(match));
ans=0;
for(int i=1;i<n;i++)
{
memset(vis,false,sizeof(vis));
if(dfs(i))
ans++;
}
}
int main()
{
int i,x,y;
while(scanf("%d",&n)==1)
{
if(n==0)
break;
scanf("%d%d",&m,&k);
for(i=1;i<=maxn;i++)
map[i].clear();
while(k--)
{
scanf("%d%d%d",&i,&x,&y);
if(x==0||y==0)
continue;
else
{
map[x].push_back(y);
}
}
hangry();
printf("%d\n",ans);
}
return 0;
}