HDU1281+二分图

一开始TLE了。。

View Code

  1 /*

  2 dfs+TLE

  3 题意：给定一些点，求最多能放多少个点，使得这些放的点不互相攻击

  4 

  5 */

  6 #include<stdio.h>

  7 #include<string.h>

  8 #include<stdlib.h>

  9 #include<algorithm>

 10 #include<iostream>

 11 #include<queue>

 12 #include<vector>

 13 #include<map>

 14 #include<math.h>

 15 typedef long long ll;

 16 //typedef __int64 int64;

 17 const int maxn = 105;

 18 const int maxm = 1005;

 19 const int inf = 0x7FFFFFFF;

 20 const double pi = acos(-1.0);

 21 const double eps = 1e-8;

 22 using namespace std;

 23 int row[ maxn ],col[ maxn ];//该行该列能否放车

 24 int ans_num;

 25 struct node2{

 26     int x,y;

 27 }Node[ 10005 ];//存储能放车的点

 28 int CNT[ maxn ][ maxn ];

 29 

 30 struct node{

 31     int x[ 24 ],y[ 24 ];

 32 }solve[ maxm ];//每个方案数的具体方法

 33 

 34 int cnt ;//记录有ansnum个车的方案数

 35 

 36 void init(){

 37     //ans_num = 0;

 38     memset( row,0,sizeof( row ) );

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

 40 }

 41 

 42 void dfs( int now,int k,int sum ){

 43     for( int i=now;i<k;i++ ){

 44         if( row[ Node[i].x ]==0&&col[ Node[i].y ]==0 ){

 45             row[ Node[i].x ] = 1;

 46             col[ Node[i].y ] = 1;

 47             if( sum+1>ans_num ) ans_num = sum+1;

 48             //printf("sum1 = %d\n",sum);

 49             dfs( now+1,k,sum+1 );

 50             row[ Node[i].x ] = 0;

 51             col[ Node[i].y ] = 0;

 52         }

 53     }

 54     return ;

 55 }

 56 

 57 void dfs2( int now,int k,int sum ){

 58     if( sum==ans_num ){

 59         //printf("sum = %d\n",sum);

 60         int f = -1;

 61         for( int j=0;j<sum;j++ ){

 62             if( solve[cnt].x[j]==-1||solve[cnt].y[j]==-1 ) {

 63                 f = 1;

 64                 break;

 65             }

 66             //CNT[ solve[cnt].x[j] ][ solve[cnt].y[j] ] ++;

 67             //printf("%d %d\n",solve[cnt].x[j],solve[cnt].y[j]);

 68         }

 69         if( f==-1 ) {

 70             

 71             //printf("sum = %d\n",sum);

 72             for( int j=0;j<sum;j++ ){

 73                 CNT[ solve[cnt].x[j] ][ solve[cnt].y[j] ] ++;

 74                 //printf("%d %d\n",solve[cnt].x[j],solve[cnt].y[j]);

 75             }

 76             

 77             cnt++ ;

 78         }

 79         return ;

 80     }

 81     for( int i=now;i<k;i++ ){

 82         if( row[ Node[i].x ]==0&&col[ Node[i].y ]==0 ){

 83             row[ Node[i].x ] = 1;

 84             col[ Node[i].y ] = 1;

 85             int tmp = cnt;

 86             solve[ tmp ].x[ sum ] = Node[i].x;

 87             solve[ tmp ].y[ sum ] = Node[i].y;

 88             //printf("sum2 = %d\n",sum+1);

 89             dfs2( now+1,k,sum+1 );

 90             solve[ tmp ].x[ sum ] = -1;

 91             solve[ tmp ].y[ sum ] = -1;

 92             row[ Node[i].x ] = 0;

 93             col[ Node[i].y ] = 0;

 94         }

 95     }

 96     return ;

 97 }

 98 

 99 int main(){

100     int n,m,k;

101     int ca = 1;

102     while( scanf("%d%d%d",&n,&m,&k)==3 ){

103         int a,b;

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

105             scanf("%d%d",&Node[i].x,&Node[i].y);

106         }

107         init();

108         ans_num = 0;

109         dfs( 0,k,0 );//求出最多能放多少个车

110         if( ans_num==0 ){

111             printf("Board %d have 0 important blanks for 0 chessmen.\n",ca++);

112             continue;

113         }

114         //printf("ans_sum:%d\n\n\n",ans_num);

115         cnt = 0;//统计放ansnum个车 有多少种方案

116         init();

117         memset( CNT,0,sizeof( CNT ) );

118         for( int i=0;i<maxm;i++ ){

119             for( int j=0;j<24;j++ )

120                 solve[i].x[j]=solve[i].y[j]=-1;

121         }

122         dfs2( 0,k,0 );

123         int res = 0;

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

125             if( CNT[Node[i].x][Node[i].y]==cnt ) res++;

126             /*printf("CNT[%d][%d] = %d\n",Node[i].x,Node[i].y,CNT[Node[i].x][Node[i].y]);*/

127         }

128         printf("Board %d have %d important blanks for %d chessmen.\n",ca++,res,ans_num);

129     }

130     return 0;

131 }

二分图：

行号和列号进行匹配！！！！！！！！！！

View Code

 1 /*

 2 二分图

 3 */

 4 #include<stdio.h>

 5 #include<string.h>

 6 #include<stdlib.h>

 7 #include<algorithm>

 8 #include<iostream>

 9 #include<queue>

10 #include<vector>

11 #include<map>

12 #include<math.h>

13 typedef long long ll;

14 //typedef __int64 int64;

15 const int maxn = 105;

16 const int maxm = 1005;

17 const int inf = 0x7FFFFFFF;

18 const double pi = acos(-1.0);

19 const double eps = 1e-8;

20 using namespace std;

21 int mat[ maxn ][ maxn ];

22 int mylink[ maxn ],vis[ maxn ];

23 

24 int ans1 ,ans2;

25 

26 bool dfs( int u,int n,int m ){

27     for( int i=1;i<=m;i++ ){

28         if( vis[ i ] ==-1&&mat[u][i]==1 ){

29             vis[ i ] = 1;

30             if( mylink[ i ]==-1||dfs( mylink[i],n,m ) ){

31                 mylink[ i ] = u;

32                 return true;

33             }

34         }

35     }

36     return false;

37 }

38 

39 int max_match( int n,int m ){

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

41     int mylink2[ maxn ];

42     int sum = 0;

43     for( int i=1;i<=n;i++ ){

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

45         if( dfs( i,n,m )==true )

46             sum++;

47     }

48     for( int i=0;i<maxn;i++ ){

49         mylink2[ i ] = mylink[ i ];

50     }

51     

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

53         if( mylink2[i]!=-1 ){

54             int tx,ty;

55             tx = mylink2[ i ];

56             ty = i;

57             mat[ tx ][ ty ] = 0;

58             memset( mylink,-1,sizeof( mylink ) );

59             int tmp_sum = 0;

60             for( int i=1;i<=n;i++ ){

61                 memset( vis,-1,sizeof( vis ));

62                 if( dfs( i,n,m )==true )

63                     tmp_sum++;

64             }

65             if( tmp_sum!=sum ) ans1++;

66             mat[ tx ][ ty ] = 1;

67         }

68     }

69     //对于已经选定的一种放车方案，分别 去掉 这种方案中的某个棋子，再看能否得到答案

70     return sum;

71 }

72  

73 int main(){

74     int n,m,k;

75     int ca = 1;

76     while( scanf("%d%d%d",&n,&m,&k)==3 ){

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

78         while( k-- ){

79             int a,b;

80             scanf("%d%d",&a,&b);

81             mat[ a ][ b ] = 1;

82         }

83         ans1 = 0;

84         ans2 = max_match( n,m );

85         printf("Board %d have %d important blanks for %d chessmen.\n",ca++,ans1,ans2);

86     }

87     return 0;

88 }