题目链接： 传送门
思路：
这道题是维基百科上面的记忆化搜索的例题。。。
四维状态dp[maxn][5][2][5]分别表示第几根棒子，这根棒子的高度，是否达到题目的要求和使用不同棒子数，那么接下来就是状态转移了。。。要用到位运算判断以前是否这种高度的棒子用到没。。。那么这个问题就解决了。。。 题目： Number of Locks
| Time Limit: 1000MS |
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Memory Limit: 10000K |
| Total Submissions: 1126 |
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Accepted: 551 |
Description In certain factory a kind of spring locks is manufactured. There are n slots (1 < n < 17, n is a natural number.) for each lock. The height of each slot may be any one of the 4 values in｛1,2,3,4｝( neglect unit ). Among the slots of a lock there are at least one pair of neighboring slots with their difference of height equal to 3 and also there are at least 3 different height values of the slots for a lock. If a batch of locks is manufactured by taking all over the 4 values for slot height and meet the two limitations above, find the number of the locks produced.
Input There is one given data n (number of slots) on every line. At the end of all the input data is -1, which means the end of input.
Output According to the input data, count the number of locks. Each output occupies one line. Its fore part is a repetition of the input data and then followed by a colon and a space. The last part of it is the number of the locks counted.
Sample Input
2
3
-1
Sample Output
2: 0
3: 8
Source Xi'an 2002
代码：
#include
#include
#include
#define New (1<<(d-1)) using namespace std; const int maxn=17+10; long long dp[maxn][5][2][5]; int n; long long dfs(int ith,int height,int k,int use,int s) { if(dp[ith][height][k][use]!=-1) return dp[ith][height][k][use]; if(ith==n) { if(k&&use>=3) return 1; else return 0; } long long ans=0; int tmp; for(int d=1;d<=4;d++) { if(!(s&New)) tmp=use+1; else tmp=use; // tmp=min(use,3); if(k||(d*height==4&&d!=2)) ans=ans+dfs(ith+1,d,1,tmp,s|New); else ans=ans+dfs(ith+1,d,0,tmp,s|New); } return dp[ith][height][k][use]=ans; } int main() { while(~scanf("%d",&n)) { if(n==-1) return -1; printf("%d: ",n); memset(dp,-1,sizeof(dp)); if(n<3) puts("0"); else { dfs(0,0,0,0,0); printf("%lld\n",dp[0][0][0][0]); } } return 0; }
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