hdu 1068 Girls and boys

Girls and Boys

Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13402 Accepted Submission(s): 6293

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

裸的二分匹配，但是我却去写最大流了2333333，真的傻，还调试了一阵子。。

借这个题学习了一下匈牙利算法。。现在写题都注重板子了23333

建两个集合，跑一遍匈牙利算法就行了

 1 #include <algorithm>
 2 #include <iostream>
 3 #include <iomanip>
 4 #include <cstring>
 5 #include <climits>
 6 #include <complex>
 7 #include <fstream>
 8 #include <cassert>
 9 #include <cstdio>
10 #include <bitset>
11 #include <vector>
12 #include <deque>
13 #include <queue>
14 #include <stack>
15 #include <ctime>
16 #include <set>
17 #include <map>
18 #include <cmath>
19 using namespace std;
20 typedef struct Edge {
21     int u, v, c, next;
22 }Edge;
23 const int inf = 0x7f7f7f7f;
24 const int maxn = 10000;
25 #define maxm 30000
26 #define ll long long
27 #define debug(x) cout<<"x="<<x<<endl
28 #define mem(a,b) memset(a,b,sizeof a)
29 inline int read()
30 {
31     int x=0,f=1;
32     char ch=getchar();
33     while(ch<'0'||ch>'9')
34     {
35         if(ch=='-') f=-1;
36         ch=getchar();
37     }
38     while(ch>='0'&&ch<='9')
39     {
40         x=10*x+ch-'0';
41         ch=getchar();
42     }
43     return x*f;
44 }
45 int  n,m,s,t,tot=0,head[maxn],l,r,match[maxn];
46 bool vis[maxn];
47 struct edge{int to,next;}e[maxm];
48 void ins(int x,int y){e[tot].to=y;e[tot].next=head[x];head[x]=tot++;}
49 bool dfs(int u)
50 {
51     for(int i=head[u];i!=-1;i=e[i].next)
52     {
53         int v=e[i].to;
54         if(vis[v]) continue;
55         vis[v]=true;
56         if(match[v]==-1||dfs(match[v]))
57         {
58             match[v]=u;
59             return true;
60         }
61     }
62     return false;
63 }
64 int hungary(int x,int y)
65 {
66     l=x;r=y;
67     int ans=0;
68     mem(match,-1);
69     for(int u=1;u<=l;++u)
70     {
71         mem(vis,0);
72         if(dfs(u)) ans++;
73     }
74     return ans;
75 }
76 void init()
77 {
78    mem(head,-1);
79    tot=0;
80 }
81 int main() {
82     int k;
83     while(~scanf("%d",&k)&&k){
84     init();
85     for(int i=0;i<k;++i)
86     {
87         int index,num,x;
88         scanf("%d: (%d) ",&index,&num);
89         for(int j=1;j<=num;++j) x=read(),
90         ins(i+1,x+1+k);
91     }
92     int hh=hungary(k,k);
93     //debug(hh);
94     hh/=2;
95     cout<<k-hh<<endl;
96     }
97     return 0;
98 }