Hdu 1068 最小路径覆盖

Girls and Boys

Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6942    Accepted Submission(s): 3118

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

最小路径覆盖　＝　顶点数　－　最大匹配数

本题由于左边集合与右边集合相同，所以每个匹配计算了两次，所以求得的最大匹配应除以２

Accepted Code:

 /*************************************************************************
     > File Name: 1068.cpp
     > Author: Stomach_ache
     > Mail: sudaweitong@gmail.com
     > Created Time: 2014年07月14日 星期一 16时29分38秒
     > Propose:
  ************************************************************************/
 
 #include <cmath>
 #include <string>
 #include <cstdio>
 #include <vector>
 #include <fstream>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 using namespace std;
 
 const int MAX_N =  + ;
 int n;
 int cx[MAX_N], cy[MAX_N], mark[MAX_N];
 vector<int> G[MAX_N];
 
 int
 path(int u) {
       for (int i = ; i < (int)G[u].size(); i++) {
           int v = G[u][i];
           if (!mark[v]) {
               mark[v] = ;
               if (cy[v] == - || path(cy[v])) {
                   cx[u] = v;
                 cy[v] = u;
                 return ;
             }
         }
     }
     return ;
 }
 
 int
 MaxMatch() {
       memset(cx, -, sizeof(cx));
     memset(cy, -, sizeof(cy));
     int res = ;
     for (int i = ; i < n; i++) {
           if (cx[i] == -) {
               memset(mark, , sizeof(mark));
             res += path(i);
         }
     }
     return res;
 }
 
 int
 main(void) {
       while (~scanf("%d", &n)) {
           for (int i = ; i < n; i++) G[i].clear();
           for (int i = ; i < n; i++) {
               int a, b;
             scanf("%d: (%d)", &a, &b);
             for (int j = ; j < b; j++) {
                   int c;
                   scanf("%d", &c);
                 G[a].push_back(c);
             }
         }
         printf("%d\n", n - MaxMatch() / );
     }
 
     return ;
 }