HDOJ---1068 Girls and Boys[匈牙利算法]

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原文链接：http://www.cnblogs.com/XBWer/archive/2012/08/06/2625392.html

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#java

本文探讨了大学二年级学生之间的浪漫关系研究，定义了一对一的浪漫关系，并寻找满足条件的最大集合，即集合中任意两人之间不存在浪漫关系。通过文本格式输入学生数据，程序输出符合条件的最大集合大小。

Girls and Boys

Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4471 Accepted Submission(s): 1944

Problem 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.

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

Source

Southeastern Europe 2000

模版：

 1 bool 寻找从k出发的对应项出的可增广路
 2 {
 3     while (从邻接表中列举k能关联到顶点j)
 4     {
 5         if (j不在增广路上)
 6         {
 7             把j加入增广路;
 8             if (j是未盖点 或者 从j的对应项出发有可增广路)
 9             {
10                 修改j的对应项为k;
11                 则从k的对应项出有可增广路,返回true;
12             }
13         }
14     }
15     则从k的对应项出没有可增广路,返回false;
16 }
17 void 匈牙利hungary()
18 {
19     for i->1 to n
20     {
21         if (则从i的对应项出有可增广路)
22             匹配数++;
23     }
24     输出 匹配数;
25 }

匈牙利算法，模版

code:

 1 #include <iostream>   
 2 #include <iomanip>   
 3 #include <fstream>   
 4 #include <sstream>   
 5 #include <algorithm>   
 6 #include <string>   
 7 #include <set>   
 8 #include <utility>   
 9 #include <queue>   
10 #include <stack>   
11 #include <list>   
12 #include <vector>   
13 #include <cstdio>   
14 #include <cstdlib>   
15 #include <cstring>   
16 #include <cmath>   
17 #include <ctime>   
18 #include <ctype.h> 
19 using namespace std;
20 
21 #define MAXN 1002
22 
23 int data[MAXN][MAXN];
24 int vst[MAXN];
25 int path[MAXN];
26 int n;
27 
28 int dfs(int v)
29 {
30     int i,j;
31     for(i=0;i<n;i++)
32         if(!vst[i]&&data[v][i])
33         {
34             vst[i]=1;
35             if(path[i]==-1||dfs(path[i]))
36             {
37                 path[i]=v;
38                 return true;
39             }
40         }
41     return false;
42 }
43 
44 int hungary()
45 {
46     int i;
47     int cnt=0;
48     memset(path,-1,sizeof(path));
49     for(i=0;i<n;i++)
50     {
51         memset(vst,0,sizeof(vst));
52         if(dfs(i))
53             cnt++;            
54     }
55     return cnt;
56 }
57 
58 int main()
59 {
60     int i,j;
61     int temp1,temp2,k;
62     int ans;
63     while(~scanf("%d",&n))
64     {
65         memset(data,0,sizeof(data));
66         for(i=0;i<n;i++)
67         {
68             scanf("%d: (%d)",&temp1,&k);
69             for(j=0;j<k;j++)
70             {
71                 scanf("%d",&temp2);
72                 data[temp1][temp2]=1;
73             }
74         }
75         ans=hungary();
76         printf("%d\n",n-ans/2);
77     }
78     return 0;
79 }