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 ...

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 
        
          2 #include 
         
           3 #include 
          
            4 #include 
           
             5 #include 
            
              6 #include 
             
               7 #include 
              
                8 #include 
               
                 9 #include 
                
                 10 #include 
                 
                  11 #include 
                  
                   12 #include 
                   
                    13 #include 
                    
                     14 #include 
                     
                      15 #include 
                      
                       16 #include 
                       
                        17 #include 
                        18 #include 
                         
                          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 3000026 #define ll long long27 #define debug(x) cout<<"x="<
                          
                           <
                           
                            '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

转载于:https://www.cnblogs.com/TYH-TYH/p/9437526.html

你可能感兴趣的文章