hdu - 1068 Girls and Boys (二分图最大独立集+拆点)

http://acm.hdu.edu.cn/showproblem.php?pid=1068

因为没有指定性别,所以要拆点,把i拆分i和i’ 那么U=V-M (M是最大匹配,U最大独立集,V是顶点数)

2U=2V-2M  所以 U=n-M‘/2. (没怎么看明白)  但是不这样会wa.

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cmath>
 4 #include <vector>
 5 #include <cstring>
 6 #include <string>
 7 #include <algorithm>
 8 #include <string>
 9 #include <set>
10 #include <functional>
11 #include <numeric>
12 #include <sstream>
13 #include <stack>
14 #include <map>
15 #include <queue>
16 #pragma comment(linker, "/STACK:102400000,102400000")
17 #define CL(arr, val)    memset(arr, val, sizeof(arr))
18
19 #define ll long long
20 #define inf 0x7f7f7f7f
21 #define lc l,m,rt<<1
22 #define rc m + 1,r,rt<<1|1
23 #define pi acos(-1.0)
24
25 #define L(x)    (x) << 1
26 #define R(x)    (x) << 1 | 1
27 #define MID(l, r)   (l + r) >> 1
28 #define Min(x, y)   (x) < (y) ? (x) : (y)
29 #define Max(x, y)   (x) < (y) ? (y) : (x)
30 #define E(x)        (1 << (x))
31 #define iabs(x)     (x) < 0 ? -(x) : (x)
32 #define OUT(x)  printf("%I64d\n", x)
33 #define lowbit(x)   (x)&(-x)
34 #define Read()  freopen("a.txt", "r", stdin)
35 #define Write() freopen("b.txt", "w", stdout);
36 #define maxn 1000000000
37 #define N 510
38 #define mod 1000000000
39 using namespace std;
40
41 vector<int>G[N];
42 int link[N];
43 bool vis[N];
44
45 bool dfs(int u)
46 {
47     for(int i=0;i<G[u].size();i++)
48     {
49         int v=G[u][i];
50         if(!vis[v])
51         {
52             vis[v]=true;
53             if(link[v]==-1||dfs(link[v]))
54             {
55                 link[v]=u;
56                 return true;
57             }
58         }
59     }
60     return false;
61 }
62 int main()
63 {
64     //Read();
65     int n,a,b,c;
66     while(~scanf("%d",&n))
67     {
68         for(int i=0;i<n;i++) G[i].clear();
69         for(int i=0;i<n;i++)
70         {
71             scanf("%d: (%d)",&a,&b);
72            // printf("%d %d\n",a,b);
73             for(int i=0;i<b;i++)
74             {
75                 scanf("%d",&c);
76                 G[a].push_back(c);
77             }
78         }
79         int ans=0;
80         memset(link,-1,sizeof(link));
81         for(int i=0;i<n;i++)
82         {
83             memset(vis,0,sizeof(vis));
84             if(dfs(i)) ans++;
85         }
86         printf("%d\n",n-ans/2);
87     }
88     return 0;
89 }