二分图最小覆盖

hdu1150 

最小顶点覆盖

 

hdu1498 50 years, 50 colors

最小顶点覆盖 +枚举

 1 #include <iostream>

 2 #include<cstdio>

 3 #include<cstring>

 4 #include<algorithm>

 5 #include<stdlib.h>

 6 #include<vector>

 7 #include<cmath>

 8 #include<queue>

 9 #include<set>

10 using namespace std;

11 #define N 105

12 #define LL long long

13 #define INF 0xfffffff

14 const double eps = 1e-8;

15 const double pi = acos(-1.0);

16 const double inf = ~0u>>2;

17 vector<int>ed[N];

18 struct node

19 {

20     int x,y;

21 };

22 vector<node>pa[N];

23 bool vis[2*N],f[N];

24 int mat[N<<1],o[N],co[N];

25 int dfs(int u)

26 {

27     int i;

28     for(i = 0 ;i < ed[u].size() ; i++)

29     {

30         int v = ed[u][i];

31         if(vis[v]) continue;

32         vis[v] = 1;

33         if(mat[v]==-1||dfs(mat[v]))

34         {

35             mat[v] = u;

36             return 1;

37         }

38     }

39     return 0;

40 }

41 int main()

42 {

43     int n,k,i,j;

44     while(scanf("%d%d",&n,&k)&&n&&k)

45     {

46         memset(f,0,sizeof(f));

47         for(i = 1; i <=100 ;i++)

48         pa[i].clear();

49         int m = 0 ;

50         for(i = 1; i <= n ;i++)

51             for(j = 1 ;j <= n ;j++)

52             {

53                 int x;

54                 node a;

55                 a.x = i;a.y = j;

56                 scanf("%d",&x);

57                 if(!f[x])

58                 {

59                     f[x] = 1;

60                     m++;

61                     co[m] = x;

62                 }

63                 pa[x].push_back(a);

64             }

65         int g = 0;

66         for(i = 1; i <= m; i++)

67         {

68             memset(mat,-1,sizeof(mat));

69             int c = co[i];

70             for(j = 1; j <= n; j++)

71             ed[j].clear();

72             for(j = 0 ;j < pa[c].size() ; j++)

73             {

74                 node a = pa[c][j];

75                 ed[a.x].push_back(a.y+n);

76             }

77             int cnt = 0;

78             for(j = 1; j <= n ;j++)

79             {

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

81                 if(dfs(j))

82                 cnt++;

83             }

84             if(cnt>k)

85             o[++g] = c;

86         }

87         if(g==0)

88         {

89             puts("-1");

90             continue;

91         }

92         sort(o+1,o+g+1);

93         for(i =1 ;i < g ; i++)

94         cout<<o[i]<<" ";

95         cout<<o[i]<<endl;

96     }

97     return 0;

98 }

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hdu2119  

最小顶点覆盖 裸

 1 #include <iostream>

 2 #include<cstdio>

 3 #include<cstring>

 4 #include<algorithm>

 5 #include<stdlib.h>

 6 #include<vector>

 7 #include<cmath>

 8 #include<queue>

 9 #include<set>

10 using namespace std;

11 #define N 110

12 #define LL long long

13 #define INF 0xfffffff

14 const double eps = 1e-8;

15 const double pi = acos(-1.0);

16 const double inf = ~0u>>2;

17 vector<int>ed[N];

18 bool vis[N<<1];

19 int mat[N<<1];

20 int dfs(int u)

21 {

22     int i;

23     for(i = 0 ;i < ed[u].size();  i++)

24     {

25         int v = ed[u][i];

26         if(vis[v]) continue;

27         vis[v] = 1;

28         if(mat[v]==-1||dfs(mat[v]))

29         {

30             mat[v] = u;

31             return 1;

32         }

33     }

34     return 0;

35 }

36 

37 int main()

38 {

39     int n,i,j,m;

40     while(cin>>n)

41     {

42         if(!n) break;

43         cin>>m;

44         memset(mat,-1,sizeof(mat));

45         for(i = 1 ;i <=  n;i++)

46         ed[i].clear();

47         for(i = 1 ;i <= n ;i++)

48             for(j = 1 ;j <= m ;j++)

49             {

50                 int x;

51                 cin>>x;

52                 if(x)

53                 ed[i].push_back(j+n);

54             }

55         int ans = 0;

56         for(i = 1 ;i <= n;i++)

57         {

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

59             if(dfs(i))

60             ans++;

61         }

62         cout<<ans<<endl;

63     }

64     return 0;

65 }

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hdu1054  Strategic Game

最小顶点覆盖 不确定左右子集 建双向边 答案/2

 1 #include <iostream>

 2 #include<cstdio>

 3 #include<cstring>

 4 #include<algorithm>

 5 #include<stdlib.h>

 6 #include<vector>

 7 #include<cmath>

 8 #include<queue>

 9 #include<set>

10 using namespace std;

11 #define N 1510

12 #define LL long long

13 #define INF 0xfffffff

14 const double eps = 1e-8;

15 const double pi = acos(-1.0);

16 const double inf = ~0u>>2;

17 int mat[N];

18 bool vis[N];

19 vector<int>ed[N];

20 int dfs(int u)

21 {

22     int i;

23     for(i = 0; i < ed[u].size(); i++)

24     {

25         int v = ed[u][i];

26         if(vis[v]) continue;

27         vis[v] = 1;

28         if(mat[v]==-1||dfs(mat[v]))

29         {

30             mat[v] = u;

31             return 1;

32         }

33     }

34     return 0;

35 }

36 int main()

37 {

38     int n,i;

39     while(scanf("%d",&n)!=EOF)

40     {

41         memset(mat,-1,sizeof(mat));

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

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

44         ed[i].clear();

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

46         {

47             int u,v,k;

48             scanf("%d%*c%*c%d%*c",&u,&k);

49             while(k--)

50             {

51                 scanf("%d",&v);

52                 ed[u].push_back(v);

53                 ed[v].push_back(u);

54             }

55         }

56         int ans  =0;

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

58         {

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

60             if(dfs(i))

61             ans++;

62         }

63         printf("%d\n",ans/2);

64     }

65     return 0;

66 }

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hdu1151   Air Raid

最小路径覆盖 裸

 1 #include <iostream>

 2 #include<cstdio>

 3 #include<cstring>

 4 #include<algorithm>

 5 #include<stdlib.h>

 6 #include<vector>

 7 #include<queue>

 8 #include<cmath>

 9 using namespace std;

10 #define N 1010

11 vector<int>pa[N];

12 bool fx[N],fy[N];

13 int vis[N],mat[N];

14 int find(int u)

15 {

16     int i;

17     for(i = 0 ; i < (int)pa[u].size() ; i++)

18     {

19         int v = pa[u][i];

20         if(vis[v]) continue;

21         vis[v] = 1;

22         if(mat[v]==0||find(mat[v]))

23         {

24             mat[v] = u;

25             return 1;

26         }

27     }

28     return 0;

29 }

30 int main()

31 {

32     int n,m,i,t;

33     scanf("%d",&t);

34     while(t--)

35     {

36         scanf("%d%d",&n,&m);

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

38         for(i = 1; i <= 1000 ; i++)

39         pa[i].clear();

40         for(i = 1; i <= m ;i++)

41         {

42             int v,u;

43             scanf("%d%d",&u,&v);

44             pa[u].push_back(v+n);

45         }

46         int s = 0;

47         for(i = 1; i <= n ;i++)

48         {

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

50             if(find(i))

51             s++;

52         }

53         printf("%d\n",n-s);

54     }

55     return 0;

56 }

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hdu1350 Taxi Cab Scheme

最小路径覆盖 

根据时间限制建边 注意：当前结束点与下一个开始点不同也可以建边 

 1 #include <iostream>

 2 #include<cstdio>

 3 #include<cstring>

 4 #include<algorithm>

 5 #include<stdlib.h>

 6 #include<vector>

 7 #include<cmath>

 8 #include<queue>

 9 #include<set>

10 using namespace std;

11 #define N 510

12 #define LL long long

13 #define INF 0xfffffff

14 const double eps = 1e-8;

15 const double pi = acos(-1.0);

16 const double inf = ~0u>>2;

17 vector<int>ed[N];

18 int ma[N][N],mat[N];

19 bool vis[N];

20 struct node

21 {

22     int t,u,v;

23     int u1,u2,v1,v2;

24     int d;

25 }p[N];

26 int dfs(int u)

27 {

28     int i;

29     for(i = 0; i < ed[u].size() ; i++)

30     {

31         int v = ed[u][i];

32         if(vis[v]) continue;

33         vis[v] = 1;

34         if(mat[v]==-1||dfs(mat[v]))

35         {

36             mat[v] = u;

37             return 1;

38         }

39     }

40     return 0;

41 }

42 int main()

43 {

44     int n,m,i,j;

45     int mod = 24*60;

46     cin>>n;

47     while(n--)

48     {

49         cin>>m;

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

51         memset(mat,-1,sizeof(mat));

52         for(i = 1; i <= m ;i++)

53         ed[i].clear();

54         int g = 0;

55         for(i = 1; i<= m ;i++)

56         {

57             int x,y,u1,u2,v1,v2;

58             scanf("%d%*c%d%d%d%d%d",&x,&y,&p[i].u1,&p[i].v1,&p[i].u2,&p[i].v2);

59             /*if(!ma[u1][v1])

60             {

61                 g++;

62                 ma[u1][v1] = g;

63             }

64             if(!ma[u2][v2])

65             {

66                 g++;

67                 ma[u2][v2] = g;

68             }

69             p[i].u = ma[u1][v1],p[i].v = ma[u2][v2];*/

70             p[i].d = abs(p[i].u1-p[i].u2)+abs(p[i].v1-p[i].v2);

71             p[i].t = x*60+y;

72         }

73         for(i = 1 ;i <= m ;i++)

74         {

75             for(j = i+1 ;j <= m; j++)

76             {

77                 int tt = abs(p[i].u2-p[j].u1)+abs(p[i].v2-p[j].v1);

78                 if(p[i].t+p[i].d+tt<p[j].t)

79                 {

80                     ed[i].push_back(j);

81                 }

82             }

83         }

84         int ans = 0;

85         for(i = 1;i <= m ;i++)

86         {

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

88             if(dfs(i))

89             ans++;

90         }

91         cout<<m-ans<<endl;

92     }

93     return 0;

94 }

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