Sharing Chocolate - UVa 1099 状压dp

Chocolate in its many forms is enjoyed by millions of people around the world every day. It is a truly universal candy available in virtually every country around the world.

You find that the only thing better than eating chocolate is to share it with friends. Unfortunately your friends are very picky and have different appetites: some would like more and others less of the chocolate that you offer them. You have found it increasingly difficult to determine whether their demands can be met. It is time to writte a program that solves the problem once and for all!

Your chocolate comes as a rectangular bar. The bar consists of
same-sized rectangular pieces. To share the chocolate you may break one
bar into two pieces along a division between rows or columns of the bar.
You or the may then repeatedly break the resulting
pieces in the same manner. Each of your friends insists on a getting a
single rectangular portion of the chocolate that has a specified number
of pieces. You are a little bit insistent as well: you will break up
your bar only if all of it can be distributed
to your friends, with none left over.

For exampla, Figure 9 shows one way that a chocolate bar consisting of 3 x 4 pieces
can be split into 4 parts that contain 6, 3, 2, and 1 pieces
respectively, by breanking it 3 times (This corresponds to the first
sample input.)

Input

The input consists of
multiple test cases each describing a chocolate bar to share. Each
description starts with a line containing a single integer n (1n15),
the number of parts in which the bar is supposed to be split. This is followed by a line containing two integers x and y (1x, y100),
the dimensions of the chocolate bar. The next line contains n positive integers, giving the number of pieces that are supposed
to be in each of the n parts.

The input is terminated by a line containing the integer zero.

Output

For each test case, first
display its case number. Then display whether it is possible to break
the chocolate in the desired way: display ``Yes" if it is possible, and ``No" otherwise. Follow the
format of the sample output.

Sample Input

4
3 4
6 3 2 1
2
2 3
1 5
0

Sample Output

Case 1: Yes
Case 2: No

题意：问x*y的巧克力能否分成给定的n块大小。

思路：dp[c][S]表示状压状态为S的巧克力能否以c为短边组成矩形，然后dfs。

AC代码如下：

[cpp] view plaincopy

1. #include<cstdio>
2. #include<cstring>
3. #include<algorithm>
4. using namespace std;
5. int sum[100010],num[20],vis[110][100010],dp[110][100010],pow2[20];
6. int n,t;
7. int dfs(int c,int S)
8. {
9. if(vis[c][S]==t)
10. return dp[c][S];
11. int i,j,k,l,ans=0,S1,S2;
12. vis[c][S]=t;
13. for(j=0;j<n;j++)
14. if(S==pow2[j])
15. return dp[c][S]=1;
16. l=sum[S]/c;
17. for(S1=(S-1)&S;S1>0;S1=(S1-1)&S)
18. {
19. S2=S-S1;
20. if(sum[S1]%c==0 && dfs(min(c,sum[S1]/c),S1) && dfs(min(c,sum[S2]/c),S2))
21. return dp[c][S]=1;
22. if(sum[S1]%l==0 && dfs(min(l,sum[S1]/l),S1) && dfs(min(l,sum[S2]/l),S2))
23. return dp[c][S]=1;
24. }
25. return dp[c][S]=0;
26. }
27. int main()
28. {
29. int i,j,k,c,l,S,ans;
30. pow2[0]=1;
31. for(i=1;i<=20;i++)
32. pow2[i]=pow2[i-1]*2;
33. while(~scanf("%d",&n) && n>0)
34. {
35. t++;
36. scanf("%d%d",&c,&l);
37. for(i=0;i<n;i++)
38. scanf("%d",&num[i]);
39. S=pow2[n]-1;
40. for(i=1;i<=S;i++)
41. {
42. sum[i]=0;
43. for(j=0;j<n;j++)
44. if(i&pow2[j])
45. sum[i]+=num[j];
46. }
47. if(c*l!=sum[S])
48. ans=0;
49. else
50. ans=dfs(min(c,l),S);
51. printf("Case %d: ",t);
52. if(ans)
53. printf("Yes\n");
54. else
55. printf("No\n");
56. }
57. }