Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13335 Accepted Submission(s): 6262
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
PS：题意就是学校里现在有n个学生，其中有些之间有暧昧关系，求最大男女没有关系的个数，这个题并没有给出那个是男的，哪个是女的，所以最后求出最大匹配数要除以2。最大独立集=顶点数-最大匹配数。
AC代码：
#include <iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<string>
#include<map>
#include<cmath>
#include<vector>
const int maxn=1e3+5;
const int mod=1e9+7;
const int inf=1e9;
#define me(a,b) memset(a,b,sizeof(a))
typedef long long ll;
using namespace std;
int n,d[maxn];
bool vis[maxn];
vector<int>maps[maxn];
int found(int x)
{
for(int i=0;i<maps[x].size();i++)
{
int u=maps[x][i];
if(!vis[u])
{
vis[u]=1;
if(d[u]==-1||found(d[u]))
{
d[u]=x;
return 1;
}
}
}
return 0;
}
int hungry()
{
me(d,-1);int sum=0;
for(int i=0;i<n;i++)
{
me(vis,0);
sum+=found(i);
}
return sum;
}
int main()
{
while(~scanf("%d",&n))
{
me(maps,0);
for(int i=0;i<n;i++)
{
int x,m;
scanf("%d: (%d)",&x,&m);
while(m--)
{
int y;scanf("%d",&y);
maps[x].push_back(y);
maps[y].push_back(x);
}
}
cout<<n-hungry()/2<<endl;
}
return 0;
}