hdu 4185 二分图匹配

题意用1*2的木板覆盖矩阵中的‘#’，（木板要覆盖的只能是‘#’），问最多能用几个木板覆盖

将#抽象为二分图的点，一个木板就是一个匹配，注意最后结果要除以2

Sample Input

1

6

......

.##...

.##...

....#.

....##

......

Sample Output

Case 1: 3

 

 1 #include<cstdio>

 2 #include<iostream>

 3 #include<algorithm>

 4 #include<cstring>

 5 #include<cmath>

 6 #include<queue>

 7 #include<map>

 8 using namespace std;

 9 #define MOD 1000000007

10 const int INF=0x3f3f3f3f;

11 const double eps=1e-5;

12 typedef long long ll;

13 #define cl(a) memset(a,0,sizeof(a))

14 #define ts printf("*****\n");

15 const int MAXN=1005;

16 int n,m,tt;

17 char s[MAXN][MAXN];

18 int uN,vN;//u,v的数目，使用前面必须赋值

19 int g[MAXN][MAXN];//邻接矩阵

20 int linker[MAXN];

21 bool used[MAXN];

22 int num[MAXN][MAXN];

23 bool dfs(int u)

24 {

25     for(int v=0;v<vN;v++)

26     if(g[u][v]&&!used[v])

27     {

28         used[v]=true;

29         if(linker[v]==-1||dfs(linker[v]))

30         {

31             linker[v] = u;

32             return true;

33         }

34     }

35     return false;

36 }

37 int hungary()

38 {

39     int res=0;

40     memset(linker,-1,sizeof(linker));

41     for(int u=0;u<uN;u++)

42     {

43         memset(used,false,sizeof(used));

44         if(dfs(u))res++;

45     }

46     return res;

47 }

48 int main()

49 {

50     int i,j,k;

51     #ifndef ONLINE_JUDGE

52     freopen("1.in","r",stdin);

53     #endif

54     int ca=0;

55     scanf("%d",&tt);

56     while(tt--)

57     {

58         scanf("%d",&n);

59         cl(g);

60         int tot=0;

61         cl(num);

62         for(i=0;i<n;i++)

63         {

64             scanf("%s",&s[i]);

65             for(j=0;j<n;j++)

66             {

67                 if(s[i][j]=='#')    num[i][j]=tot++;

68             }

69         }

70 

71         uN=vN=tot;

72         for(i=0;i<n;i++)

73         {

74             for(j=0;j<n;j++)

75             {

76                 if(s[i][j]=='#')

77                 {

78                     if(s[i+1][j]=='#'&&i+1<n)  g[num[i][j]][num[i+1][j]]=1;

79                     if(s[i][j+1]=='#'&&j+1<n)  g[num[i][j]][num[i][j+1]]=1;

80                     if(s[i-1][j]=='#'&&i-1>=0)  g[num[i][j]][num[i-1][j]]=1;

81                     if(s[i][j-1]=='#'&&j-1>=0)  g[num[i][j]][num[i][j-1]]=1;

82                 }

83             }

84         }

85         ca++;

86         printf("Case %d: %d\n",ca,hungary()/2);

87     }

88 }