【二分圖最大獨立集】poj 1466 Girls and Boys
阿新 • • 發佈:2020-11-27
Girls and Boys
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.
For each given data set, the program should write to standard output a line containing the result.
Time Limit: 5000MS | Memory Limit: 10000K | |
Total Submissions: 14701 | Accepted: 6613 |
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 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
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
最大獨立集=點數-最大匹配數
這道題目的cnt要除以2,因為沒分二分圖的兩側
程式碼
#include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> using namespace std; int n,ma[505][505],link[505],vis[505]; int dfs(int x) { for(int i=1;i<=n;i++) { if(!vis[i] && ma[x][i]==1) { vis[i]=1; if(!link[i] || dfs(link[i])==1) { link[i]=x; return 1; } } } return 0; } int main() { freopen("a.in","r",stdin); while(scanf("%d",&n)!=EOF) { memset(ma,0,sizeof(ma)); memset(link,0,sizeof(link)); int k,x,y; for(int i=1;i<=n;i++) { scanf("%d: (%d)",&k,&x); k++; for(int j=1;j<=x;j++) { scanf("%d",&y); y++; ma[k][y]=1; } } int cnt=0; for(int i=1;i<=n;i++) { memset(vis,0,sizeof(vis)); if(dfs(i)) cnt++; } printf("%d\n",n-cnt/2); } return 0; }