ACM: 图论题 poj 1466 最大独立团

Girls and
Boys
 

Description

In 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.
Input

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 (n
<=500 ), for n subjects.
Output

For each given data set, the
program should write to standard output a line containing the
result.
Sample 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

Sample Output

5

2

 

题意: 男生和女生组合, 求出独立的人的个数.

 

解题思路:

     1.
最大独立团公式: result = n-最大匹配/2

 

代码:

#include <cstdio>

#include <iostream>

#include <cstring>

using namespace std;

#define MAX 501

int n;

int x[MAX], y[MAX];

bool g[MAX][MAX], vis[MAX];

bool find(int u)

{

 for(int v = 0; v < n; ++v)

 {

  if( g[u][v]
&& !vis[v] )

  {

   vis[v] =
true;

   if( y[v] ==
-1 || find(y[v]) )

   {

    y[v]
= u;

    return
true;

   }

  }

 }

 return false;

}

int Edmonds()

{

 int result = 0;

 memset(y, -1, sizeof(y));

 for(int u = 0; u < n; ++u)

 {

  memset(vis, false,
sizeof(vis));

  if( find(u) ) result++;

 }

 return result;

}

int main()

{

// freopen("input.txt", "r", stdin);

 while(scanf("%d", &n) !=
EOF)

 {

  int u, v, num;

  memset(g, false,
sizeof(g));

  for(int i = 0; i
< n; ++i)

  {

   scanf("%d:
(%d)", &u, &num);

   for(int j =
0; j < num; ++j)

   {

    scanf("%d",
&v);

    g[u][v]
= true;

   }

  }

  printf("%d\n",
n-Edmonds()/2);

 }

 return 0;

}