HDU 1068 Girls and Boys(二分图最大独立点集,3级)

本文介绍了一个关于大学生恋爱关系的研究问题,旨在找出没有恋爱关系的学生集合的最大数量。通过构造图论中的二分图并使用匹配算法解决该问题。
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 Girls and Boys
Time Limit:10000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

 

Appoint description:  

Description

the second year of the university somebody started a study on the romantic relations between the students. The relation “romantically involved” is defined between one girl and one boy. For the study reasons it is necessary to find out the maximum set satisfying the condition: there are no two students in the set who have been “romantically involved”. The result of the program is the number of students in such a set.

The input contains several data sets in text format. Each data set represents one set of subjects of the study, with the following description:

the number of students
the description of each student, in the following format
student_identifier:(number_of_romantic_relations) student_identifier1 student_identifier2 student_identifier3 ...
or
student_identifier:(0)

The student_identifier is an integer number between 0 and n-1, for n subjects.
For each given data set, the program should write to standard output a line containing the result.
 

Input

     
7 0: (3) 4 5 6 1: (2) 4 6 2: (0) 3: (0) 4: (2) 0 1 5: (1) 0 6: (2) 0 1 3 0: (2) 1 2 1: (1) 0 2: (1) 0
 

Output

     
5 2
 

 

思路:ans=n-二分图最大独立点集/2

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
const int mm=1e4+9;
const int mn=5e6+9;
class node
{
  public:int v,next;
}e[mn];
int head[mm],edge,n,dot;
int Left[mm];
bool vis[mm];
void data()
{
  memset(head,-1,sizeof(head));dot=edge=0;
}
void add(int u,int v)
{
  e[edge].v=v;e[edge].next=head[u];head[u]=edge++;
}
void read()
{ int u,v,m;char c;
  scanf("%d:",&u);
  if(u==-1)u=0;
  while(1)
  {
    c=getchar();
    if(c=='(')break;
  }
  scanf("%d",&m);dot=max(dot,u);
  while(1)
  {
    c=getchar();
    if(c==')')break;
  }
  for(int i=0;i<m;++i)
  {
    scanf("%d",&v);dot=max(dot,v);
    if(u<v){add(u,v);add(v,u);}
  }
}
int dfs(int u)
{ int v;
  for(int i=head[u];~i;i=e[i].next)
  {
    v=e[i].v;
    if(vis[v])continue;
    vis[v]=1;
    if(Left[v]==-1||dfs(Left[v]))
    {
      Left[v]=u;return 1;
    }
  }
  return 0;
}
int get_max()
{ int ret=0;
  memset(Left,-1,sizeof(Left));
  for(int i=0;i<=dot;++i)
  { memset(vis,0,sizeof(vis));
    if(dfs(i))
      ++ret;
  }
  return ret;
}
int main()
{ int u,v,m;char c;
  while(~scanf("%d",&n))
  { data();
    for(int i=0;i<n;++i)
    {
      read();
    }
    int ans=n-get_max()/2;
    printf("%d\n",ans);
  }
  return 0;
}


 

 

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