| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 13731 | Accepted: 5873 |
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
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
#include<cstdlib>
#include<iomanip>
using namespace std;
int read(){
int x=,f=;char ch=getchar();
while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
return x*f;
}
int n,m,k;
int nx,ny,ans=;
bool map[][];
bool ex[],ey[];
int cx[],cy[];
void change();
int dfs(int);
int main()
{
std::ios::sync_with_stdio(false);
while(cin>>n)
{
if(n==) break;
cin>>m>>k;
memset(map,false,sizeof(map));
for(int i=;i<k;i++)
{
int a,b,c;
cin>>a>>b>>c;
map[b][c]=true;
}
change();
cout<<ans<<endl;
}
return ;
}
void change()
{
ans=;
int a,b,c;
memset(cx,,sizeof(cx));memset(cy,,sizeof(cy));
for(int i=;i<=n;i++)
{
if(!cx[i])
{
memset(ex,false,sizeof(ex));memset(ey,false,sizeof(ey));
ans+=dfs(i);
}
}
return;
}
int dfs(int u)
{
int a,b,c;
ex[u]=true;
int v;
for(v=;v<=m;v++)
{
if((map[u][v]==true)&&(ey[v]==false))
{
ey[v]=true;
if(!cy[v]||dfs(cy[v]))
{
cx[u]=v;cy[v]=u;
return ;
}
}
}
return ;
}
POJ 1325