HDU1507 Uncle Tom's Inherited Land*

题目是跟 zoj1516是一样的，但多了匹配后的输出

详解zoj1516可见http://www.cnblogs.com/CSU3901130321/p/4228057.html

 1 #include <cstdio>

 2 #include <cstring>

 3 #include <cmath>

 4 using namespace std;

 5 

 6 const int maxn = 55;

 7 int n , m , k;

 8 int g[maxn][maxn] , cx[maxn] , cy[maxn] , visy[maxn] , nx , ny;

 9 bool col[105][105];//判断是否为鱼塘

10 

11 struct Point{

12     int x , y;

13     Point(int x=0 , int y=0):x(x),y(y){}

14 };

15 

16 Point px[maxn] , py[maxn];

17 

18 bool ok(Point p1 , Point p2)

19 {

20     if((p1.x == p2.x && abs(p1.y - p2.y) == 1) || (p1.y == p2.y && abs(p1.x - p2.x) == 1))

21         return true;

22     return false;

23 }

24 

25 void build_graph()

26 {

27     memset(g , 0 , sizeof(g));

28     for(int i=1 ; i<=nx ; i++){

29         for(int j=1 ; j<=ny ; j++){

30             if(ok(px[i] , py[j]))

31                 g[i][j] = 1;

32         }

33     }

34 }

35 

36 int dfs(int u)

37 {

38     for(int v = 1 ; v<=ny ; v++)

39         if(g[u][v] && !visy[v]){

40             visy[v] = 1;

41             if(cy[v] == -1 || dfs(cy[v])){

42                 cx[u] = v;

43                 cy[v] = u;

44                 return 1;

45             }

46         }

47     return 0;

48 }

49 

50 int MaxMatch()

51 {

52     memset(cx , -1 , sizeof(cx));

53     memset(cy , -1 , sizeof(cy));

54     int ans = 0;

55     for(int i=1 ; i<=nx ; i++){

56         if(cx[i] == -1){

57             memset(visy , 0 , sizeof(visy));

58             ans += dfs(i);

59         }

60     }

61     return ans;

62 }

63 

64 int main()

65 {

66   //  freopen("a.in" , "r"  ,stdin);

67     while(scanf("%d%d" , &n , &m) , n)

68     {

69         scanf("%d" , &k);

70         int x , y;

71         memset(col , 0 , sizeof(col));

72         while(k--){

73             scanf("%d%d" , &x , &y);

74             col[x][y] = 1;

75         }

76         nx = 0 , ny = 0;

77         for(int i=1 ; i<=n ; i++)

78             for(int j=1 ; j<=m ; j++)

79                 if(!col[i][j]){

80                     if((i+j)&1) py[++ny] = Point(i,j);

81                     else px[++nx] = Point(i,j);

82                 }

83         //构造二分图

84         build_graph();

85         printf("%d\n" , MaxMatch());

86         for(int i=1 ; i<=nx ; i++){

87             if(cx[i] != -1)

88                 printf("(%d,%d)--(%d,%d)\n",px[i].x , px[i].y , py[cx[i]].x , py[cx[i]].y);

89         }

90     }

91     return 0;

92 }