POJ1446 Girls and Boys
Time Limit: 5000MS | Memory Limit: 10000K | |
Total Submissions: 12747 | Accepted: 5678 |
Description
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
Source
——————————————————————————————————
题目的意思是给出n个学生喜欢的关系,问最多选出多少个人没有喜欢关系
思路:求最大独立集,最大独立集=点数n-最大匹配数
#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <vector>
#include <set>
#include <stack>
#include <map>
#include <climits> using namespace std; #define LL long long
const int INF = 0x3f3f3f3f; const int MAXN=1000;
int uN,vN; //u,v数目
int g[MAXN][MAXN];//编号是0~n-1的
int linker[MAXN];
bool used[MAXN];
int mat[MAXN];
int aa[MAXN];
struct area
{
int x1,x2,y1,y2;
} s[100005];
struct point
{
int x,y;
} p[100005]; bool dfs(int u)
{
int v;
for(v=0; v<vN; v++)
if(g[u][v]&&!used[v])
{
used[v]=true;
if(linker[v]==-1||dfs(linker[v]))
{
linker[v]=u;
return true;
}
}
return false;
}
int hungary()
{
int res=0;
int u;
memset(linker,-1,sizeof(linker));
for(u=0; u<uN; u++)
{
memset(used,0,sizeof(used));
if(dfs(u)) res++;
}
return res;
} int main()
{
int n,m,x,y;
int cas=1;
while(~scanf("%d",&n))
{
memset(g,0,sizeof g);
for(int i=0; i<n; i++)
{
scanf("%d: (%d)",&x,&m);
for(int i=0; i<m; i++)
{
scanf("%d",&y);
g[x][y]=1;
}
}
uN=vN=n;
printf("%d\n",n-hungary()/2);
}
return 0;
}
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