Poj2594 treasure exploration [bipartite graph minimum path coverage] [Floyd]

Topic link ：

​ ​http://poj.org/problem?id=2594​​

The main idea of the topic ：

Here you are. N Locations ,M The strip has a directed edge , A graph of known composition is a directed acyclic graph . Now let the robot pass on the site M

Edge to traverse N Locations , ask ： At least how many robots are needed to traverse N Locations .

Ideas ：

This is a problem of finding the minimum path coverage . What is different from the problem of general minimum path coverage is ： The points here are

To iterate . That is, two or more robots can pass through the same point . that , First establish a bipartite graph ,

On both sides N Locations . Then based on the original drawing , use Floyd Find a transitive closure , That is, if you point i It can reach

spot j, And point j Can reach the point k, Then it can be regarded as a point i You can skip the point directly j And the point of arrival k, You can establish a directed

edge (i,k). After building the map , Is the general bipartite graph minimum path coverage problem . And bipartite graph minimum path coverage = 

points - Maximum matching of bipartite graph , Use Hungarian algorithm to find the maximum match of bipartite graph .

AC Code ：

Log in and copy

      
      

       
       #include<iostream>
       
       

       
       #include<algorithm>
       
       

       
       #include<cstdio>
       
       

       
       #include<cstring>
       
       

       
       using 
       
       namespace 
       
       std;
       
       

       
       const 
       
       int 
       
       MAXN 
       
       = 
       
       550;
       
       

       
       

       
       bool 
       
       Map[
       
       MAXN][
       
       MAXN],
       
       Mask[
       
       MAXN];
       
       

       
       int 
       
       NX,
       
       NY;
       
       

       
       int 
       
       cx[
       
       MAXN],
       
       cy[
       
       MAXN];
       
       

       
       

       
       void 
       
       Floyd(
       
       int 
       
       N)       
       
       // Finding transitive closure 
       
       
{
       
       
    
       
       for(
       
       int 
       
       k 
       
       = 
       
       1; 
       
       k 
       
       <= 
       
       N; 
       
       ++
       
       k)
       
       
    {
       
       
        
       
       for(
       
       int 
       
       i 
       
       = 
       
       1; 
       
       i 
       
       <= 
       
       N; 
       
       ++
       
       i)
       
       
        {
       
       
            
       
       for(
       
       int 
       
       j 
       
       = 
       
       1; 
       
       j 
       
       <= 
       
       N; 
       
       ++
       
       j)
       
       
            {
       
       
                
       
       if(
       
       i 
       
       != 
       
       j 
       
       && 
       
       Map[
       
       i][
       
       k] 
       
       && 
       
       Map[
       
       k][
       
       j])
       
       
                    
       
       Map[
       
       i][
       
       j] 
       
       = 
       
       1;
       
       
            }
       
       
        }
       
       
    }
       
       
}
       
       

       
       

       
       int 
       
       FindPath(
       
       int 
       
       u)
       
       
{
       
       
    
       
       for(
       
       int 
       
       i 
       
       = 
       
       1; 
       
       i 
       
       <= 
       
       NY; 
       
       ++
       
       i)
       
       
    {
       
       
        
       
       if(
       
       Map[
       
       u][
       
       i] 
       
       && 
       
       !
       
       Mask[
       
       i])
       
       
        {
       
       
            
       
       Mask[
       
       i] 
       
       = 
       
       1;
       
       
            
       
       if(
       
       cy[
       
       i] 
       
       == 
       
       -
       
       1 
       
       || 
       
       FindPath(
       
       cy[
       
       i]))
       
       
            {
       
       
                
       
       cy[
       
       i] 
       
       = 
       
       u;
       
       
                
       
       cx[
       
       u] 
       
       = 
       
       i;
       
       
                
       
       return 
       
       1;
       
       
            }
       
       
        }
       
       
    }
       
       
    
       
       return 
       
       0;
       
       
}
       
       

       
       

       
       int 
       
       MaxMatch()  
       
       // Maximum matching of bipartite graph 
       
       
{
       
       
    
       
       for(
       
       int 
       
       i 
       
       = 
       
       1; 
       
       i 
       
       <= 
       
       NX; 
       
       ++
       
       i)
       
       
        
       
       cx[
       
       i] 
       
       = 
       
       -
       
       1;
       
       
    
       
       for(
       
       int 
       
       i 
       
       = 
       
       1; 
       
       i 
       
       <= 
       
       NY; 
       
       ++
       
       i)
       
       
        
       
       cy[
       
       i] 
       
       = 
       
       -
       
       1;
       
       

       
       
    
       
       int 
       
       res 
       
       = 
       
       0;
       
       
    
       
       for(
       
       int 
       
       i 
       
       = 
       
       1; 
       
       i 
       
       <= 
       
       NX; 
       
       ++
       
       i)
       
       
    {
       
       
        
       
       if(
       
       cx[
       
       i] 
       
       == 
       
       -
       
       1)
       
       
        {
       
       
            
       
       for(
       
       int 
       
       j 
       
       = 
       
       1; 
       
       j 
       
       <= 
       
       NY; 
       
       ++
       
       j)
       
       
                
       
       Mask[
       
       j] 
       
       = 
       
       0;
       
       
            
       
       res 
       
       += 
       
       FindPath(
       
       i);
       
       
        }
       
       
    }
       
       
    
       
       return 
       
       res;
       
       
}
       
       

       
       

       
       int 
       
       main()
       
       
{
       
       
    
       
       int 
       
       N,
       
       M,
       
       u,
       
       v;
       
       
    
       
       while(
       
       ~scanf(
       
       "%d%d",
       
       &
       
       N,
       
       &
       
       M) 
       
       && (
       
       N
       
       ||
       
       M))
       
       
    {
       
       
        
       
       memset(
       
       Map,
       
       0,
       
       sizeof(
       
       Map));
       
       
        
       
       for(
       
       int 
       
       i 
       
       = 
       
       1; 
       
       i 
       
       <= 
       
       M; 
       
       ++
       
       i)
       
       
        {
       
       
            
       
       scanf(
       
       "%d%d",
       
       &
       
       u,
       
       &
       
       v);
       
       
            
       
       Map[
       
       u][
       
       v] 
       
       = 
       
       1;
       
       
        }
       
       
        
       
       Floyd(
       
       N);
       
       
        
       
       NX 
       
       = 
       
       NY 
       
       = 
       
       N;
       
       
        
       
       printf(
       
       "%d\n",
       
       N 
       
       - 
       
       MaxMatch());
       
       
    }
       
       

       
       
    
       
       return 
       
       0;
       
       
}
      
      
      
      

       
       
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