POJ2446 二分图最大匹配

问题：POJ2446

分析：

　　采用黑白相间的方法把棋盘分成两个点集，是否可以用1*2的卡片实现全覆盖等价于二分图是否有完全匹配。

AC代码

 1 //Memory: 172K        Time: 32MS

 2 #include <iostream>

 3 #include <cstring>

 4 #include <cstdio>

 5 

 6 using namespace std;

 7 

 8 const int maxn =  32 * 32 / 2 + 10;

 9 int maze[33][33];

10 int edge[maxn][4];

11 int ne[maxn];

12 int match[maxn];

13 int vis[maxn];

14 int m, n, k;

15 int x, y;

16 int flag;

17 int num1, num0;

18 const int s[4][2] = { {1, 0}, {-1, 0}, {0, 1}, {0, -1}};

19 

20 

21 bool findPath(int start)

22 {

23     for (int i = 0; i < ne[start]; i++){

24         int current = edge[start][i];

25         if ( vis[current] ) continue;

26         vis[current] = 1;

27         if ( !match[current] || findPath( match[current] ) ){

28             match[current] = start;

29             return true;

30         }

31     }

32     return false;

33 }

34 

35 bool solve()

36 {

37     memset(match, 0, sizeof(match));

38     int cnt = 0;

39     for (int i = 1; i <= num0; i++){

40         memset(vis, 0, sizeof(vis));

41         if ( findPath(i) )

42             ++cnt;

43     }

44     if (cnt == num0) return true;

45     return false;

46 }

47 

48 void input()

49 {

50     memset(edge, 0, sizeof(edge));

51     memset(ne, 0, sizeof(ne));

52     memset(maze, 0, sizeof(maze));

53     scanf("%d%d%d", &m, &n, &k);

54     for (int i = 0; i < k; i++){

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

56         maze[y - 1][x - 1] =  -1;

57     }

58     num0 = num1 = 0;

59     for (int i = 0; i < m; i++)

60         for (int j = 0; j < n; j++){

61             if ( maze[i][j] != -1 && (i + j) % 2 == 0) {

62                 maze[i][j] = ++num0;

63             }

64             else if (maze[i][j] != -1 && (i + j) % 2 == 1) {

65                 maze[i][j] = ++num1;

66             }

67         }

68     if ( num0 != num1 ) return;

69     for (int i = 0; i < m; i++) {

70         for (int j = 0; j < n; j++){

71             if ( (i + j) % 2 == 0 && maze[i][j] != -1){

72                 int current = maze[i][j];

73                 for (int k = 0; k < 4; k++){

74                     int ni = i + s[k][0];

75                     int nj = j + s[k][1];

76                     if (ni >= m || ni < 0 || nj >= n || nj < 0)

77                         continue;

78                     if ( maze[ni][nj] != -1) edge[current][ne[current]++] = maze[ni][nj];

79                 }

80             }

81         }

82     }

83 }

84 

85 int main()

86 {

87     input();

88     if ( num0 == num1 && solve() ) printf("YES\n");

89     else printf("NO\n");

90     return 0;

91 }