hdu 3829 最大独立集

思路：很显然，让某一个孩子happy是一定可以做到的，但是同时有可能会让别的孩子unhappy，所以每一对这样的两个孩子之间存在”冲突”，如果在存在“冲突”的孩子之间建边的话，我们的问题就转化成了求二分图的最大独立集。具体为什么是二分图，博主也没有想明白...

 1 #include <iostream>
 2 #include <cstring>
 3 #include <cstdio>
 4 using namespace std;
 5
 6 const int N = 501;
 7 const int M = N * N;
 8 const int L = 11;
 9 bool visit[N];
10 int head[N];
11 int mark[N];
12 int n, m, p, e;
13
14 struct Edge
15 {
16     int v, next;
17 } edge[M];
18
19 struct Hobby
20 {
21     char like[L], dislike[L];
22 } hobby[N];
23
24 void init()
25 {
26     e = 0;
27     memset( head, -1, sizeof(head) );
28 }
29
30 void addEdge( int u, int v )
31 {
32     edge[e].v = v;
33     edge[e].next = head[u];
34     head[u] = e++;
35 }
36
37 int dfs( int u )
38 {
39     for ( int i = head[u]; i != -1; i = edge[i].next )
40     {
41         int v = edge[i].v;
42         if ( !visit[v] )
43         {
44             visit[v] = true;
45             if ( mark[v] == -1 || dfs( mark[v] ) )
46             {
47                 mark[v] = u;
48                 mark[u] = v;
49                 return 1;
50             }
51         }
52     }
53     return 0;
54 }
55
56 int hunagry()
57 {
58     memset( mark, -1, sizeof(mark) );
59     int res = p;
60     for ( int i = 1; i <= p; i++ )
61     {
62         if ( mark[i] != -1 ) continue;
63         memset( visit, 0, sizeof(visit) );
64         res -= dfs(i);
65     }
66     return res;
67 }
68
69 int main()
70 {
71     while ( scanf("%d%d%d", &n, &m, &p) != EOF )
72     {
73         init();
74         for ( int i = 1; i <= p; i++ )
75         {
76             scanf("%s%s", hobby[i].like, hobby[i].dislike);
77             for ( int j = i - 1; j > 0; j-- )
78             {
79                 if ( strcmp( hobby[i].like, hobby[j].dislike ) == 0
80                 ||   strcmp( hobby[j].like, hobby[i].dislike ) == 0 )
81                 {
82                     addEdge( i, j );
83                     addEdge( j, i );
84                 }
85             }
86         }
87         printf("%d\n", hunagry());
88     }
89     return 0;
90 }