POJ 2441 Arrange the BUlls 状压DP

Arrange the Bulls

Time Limit: 4000MS		Memory Limit: 65536K
Total Submissions: 3752		Accepted: 1438

Description

Farmer Johnson's Bulls love playing basketball very much. But none of them would like to play basketball with the other bulls because they believe that the others are all very weak. Farmer Johnson has N cows (we number the cows from 1 to N) and M barns (we number the barns from 1 to M), which is his bulls' basketball fields. However, his bulls are all very captious, they only like to play in some specific barns, and don’t want to share a barn with the others. 

So it is difficult for Farmer Johnson to arrange his bulls, he wants you to help him. Of course, find one solution is easy, but your task is to find how many solutions there are. 

You should know that a solution is a situation that every bull can play basketball in a barn he likes and no two bulls share a barn. 

To make the problem a little easy, it is assumed that the number of solutions will not exceed 10000000.

Input

In the first line of input contains two integers N and M (1 <= N <= 20, 1 <= M <= 20). Then come N lines. The i-th line first contains an integer P (1 <= P <= M) referring to the number of barns cow i likes to play in. Then follow P integers, which give the number of there P barns.

Output

Print a single integer in a line, which is the number of solutions.

Sample Input

3 4

2 1 4

2 1 3

2 2 4

Sample Output

4

Source

POJ Monthly,TN

 

 

 

 

 

题意：给出n只球队和m个篮球场，要把这m个篮球场分配给这n只球队，要求，每个篮球场只能分配给一只球队。

而且每只球队都要指定的若干个篮球场中的一个。

问分配可方案个数。

 

状压DP，下标j的二进制表示，1表示这个篮球场还没有分配，0表示这个篮球场已经被分配了。

 

 

dp[i][j]表示前i只球队，在j的二进制表示这个球场的状态下的方案个数。

 

刚开始mle了，改为滚动数组

然后，tle了，又加了一句。立马2500ms过了。

 

注意：DP的优化很大一个方面就是对决策个数的优化，在循环的过程中，尽量减少对不是最优决策的更新。

 

还可以对转移方程自身进行转换，化简，达到优化的目的。

 

 

 

 

 

 1 #include<cstdio>

 2 #include<cstring>

 3 #include<algorithm>

 4 #include<vector>

 5 

 6 using namespace std;

 7 

 8 const int maxn=20;

 9 

10 int dp[2][1<<maxn];

11 int sum[1<<maxn];

12 vector<int>a[maxn];

13 

14 int main()

15 {

16     for(int i=0;i<(1<<20);i++)

17     {

18         sum[i]=0;

19         for(int j=0;j<20;j++)

20             if(i&(1<<j))

21                 sum[i]+=1;

22     }

23     int n,m;

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

25     {

26         for(int i=0;i<=n;i++)

27             a[i].clear();

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

29         {

30             int k;

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

32             while(k--)

33             {

34                 int u;

35                 scanf("%d",&u);

36                 a[i].push_back(u);

37             }

38         }

39         for(int i=0;i<2;i++)

40         {

41             for(int j=0;j<(1<<m);j++)

42                 dp[i][j]=0;

43         }

44         dp[0][(1<<m)-1]=1;

45 

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

47         {

48             for(int j=(1<<m)-1;j>=0;j--)

49             {

50                 if(dp[(i-1)%2][j]==0)       //不加这句会tle

51                     continue;

52                 for(int k=0;k!=a[i].size();k++)

53                 {

54                     int tmp=a[i][k];

55                     if(!(j&(1<<(tmp-1))))

56                         continue;

57                     int x=j-(1<<(tmp-1));

58                     dp[i%2][x]+=dp[(i-1)%2][j];

59                 }

60             }

61         }

62 

63         int ans=0;

64         for(int i=0;i<(1<<m);i++)

65         {

66             if(m-sum[i]>=n)

67             {

68                 ans+=dp[n%2][i];

69             }

70         }

71         printf("%d\n",ans);

72     }

73     return 0;

74 }

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