二分匹配的Hopcroft-Carp算法

HDU 1150 Machine Schedule问题

Machine Schedule

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4062    Accepted Submission(s): 1976

Problem 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
   
   
   
   

Hopcroft-Carp算法快，用匈牙利算法15ms，而Hopcorft-Carp却0ms。因为匈牙利算法的时间复杂度为O(n*e)，而Hopcroft-Carp算法O(sqrt(n)*e)

本算法适合处理大一点的数据.

代码：

/*Hopcroft-Carp算法
这个算法比匈牙利算法的时间复杂度要小，大数据可以采用这个算法
二分图匹配（Hopcroft-Carp的算法）。
初始化：g[][]邻接矩阵
调用：res=MaxMatch();  Nx,Ny要初始化！！！
时间复杂大为 O(V^0.5 E)
适用于数据较大的二分匹配
需要queue头文件
*/
#include<iostream>
#include<stdio.h>
#include<memory>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
const  int MAXN=1000;
const int INF=1<<28;
int g[MAXN][MAXN],Mx[MAXN],My[MAXN],Nx,Ny;
int dx[MAXN],dy[MAXN],dis;
bool visit[MAXN];

bool searchP()
{
 queue<int> q;
 dis=INF;
 memset(dx,-1,sizeof(dx));
 memset(dy,-1,sizeof(dy));
 for(int i=0;i<Nx;i++)
 if(Mx[i]==-1)
  {
   q.push(i);
   dx[i]=0;
  }
while(!q.empty())
{
 int u=q.front();
 q.pop();
 if(dx[u]>dis) break;
 for(int v=0;v<Ny;v++)
  if(g[u][v]&&dy[v]==-1)
  {
   dy[v]=dx[u]+1;
   if(My[v]==-1) dis=dy[v];
   else
   {
    dx[My[v]]=dy[v]+1;
    q.push(My[v]);
   }
  }
 }
 return dis!=INF;
}
bool DFS(int u)
{
    for(int v=0;v<Ny;v++)
    if(!visit[v]&&g[u][v]&&dy[v]==dx[u]+1)
    {
          visit[v]=1;
          if(My[v]!=-1&&dy[v]==dis) continue;
          if(My[v]==-1||DFS(My[v]))
          {
               My[v]=u;
               Mx[u]=v;
               return 1;
          }

       }
  return 0;
}

int MaxMatch()
{
int res=0;
memset(Mx,-1,sizeof(Mx));
memset(My,-1,sizeof(My));
while(searchP())
{
memset(visit,0,sizeof(visit));
for(int i=0;i<Nx;i++)
 if(Mx[i]==-1&&DFS(i)) res++;
}
return res;
}
int main()
{
    int k,i,u,v;
    while(scanf("%d",&Nx),Nx)
    {
     scanf("%d%d",&Ny,&k);
     memset(g,0,sizeof(g));
     while(k-- >0)
     {
      scanf("%d%d%d",&i,&u,&v);
      if(u>0&&v>0) g[u][v]=1;//初始状态为0，一开始0的边不要加
     }
    printf("%d\n",MaxMatch());
    }
    return 0;
}