Poj 1466 (maximum vertex independence set)

By the Konig theorem, the minimum vertex overwrite = The maximum match

It can be found that in a bipartite graph, as long as the vertices in the minimum vertices overwrite set are removed, there is no edge between the remaining vertices, that is, the remaining vertices are the largest vertices independent set.

Then you can use (maximum vertex independence set) = (n-least vertex overwrite) = (n-maximum match)

 

Girls and boys Time limit: 5000 Ms   Memory limit: 10000 K Total submissions: 9325   Accepted: 4123 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 shocould write to standard output a line containing the result. Sample Input 70: (3) 4 5 61: (2) 4 62: (0) 3: (0) 4: (2) 0 15: (1) 06: (2) 0 0 130: (2) 1 21: (1) 02: (1) 0 Sample output 52 Source Southeastern Europe 2000

 

# Include <stdio. h> # Include < String . H> # Define N 1010 Int  N;  Int G [ 505 ] [ 505  ];  Int  CNT, pre [N];  Int  Mark [N], FRT [N];  Int  Save1 [N], save2 [N];  Int  Cnt1, cnt2;  Void Fdfs ( Int S, Int  F ){  If (F = 0 ) Save1 [cnt1 ++] = S;  Else Save2 [cnt2 ++] = S; Mark [s] = F;  For ( Int I = 0 ; I <n; I ++ ){  If (G [s] [I] = 0 ) Continue  ;  If (MARK [I] =- 1 ) Fdfs (I, f ^ 1  );}}  Int DFS ( Int  S ){  For ( Int I = 0 ; I <cnt2; I ++ ){  If (G [save1 [s] [save2 [I] = 0 | Mark [I] = 1 )Continue  ; Mark [I] = 1  ;  If (FRT [I] =- 1 | DFS (FRT [I]) {FRT [I] = S;  Return   1  ;}}  Return   0  ;}  Int Main (){  Int  Ans;  While (Scanf ( "  % D  " , & N )! = EOF) {cnt1 = Cnt2 = 0  ; Ans = 0  ; Memset (G,  0 , Sizeof  (G )); For ( Int I = 0 ; I <n; I ++ ){  Int  X, Y; scanf (  "  % D: (% d)  " , & X ,& Y );  //  If (y = 0) ans ++;              For ( Int J = 0 ; J <Y; j ++ ){  Int  TMP; scanf (  "  % D  " ,& TMP); G [I] [TMP] = G [TMP] [I] = 1  ;} Memset (mark, - 1 , Sizeof  (Mark ));  For (Int I = 0 ; I <n; I ++ ){  If (MARK [I]! =- 1 ) Continue  ; Fdfs (I,  0  );}  Int Sum = 0  ; Memset (FRT, - 1 , Sizeof (FRT ));  For ( Int I = 0 ; I <cnt1; I ++ ) {Memset (mark,  0 , Sizeof  (Mark); sum + = DFS (I);} printf (  "  % D \ n  " , N- Sum );}  Return  0  ;}