POJ 1351-Number of Locks（记忆化搜索） - c++编程基础

Number of Locks

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 1140		Accepted: 559

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

4维的记忆化搜索。。英语太渣。。看的题解。

题意：有一排n个柱子。每个柱子的高度可以为{1,2,3,4} 求问 排列中存在相邻两个柱子高度差为3 ，且使用了至少3种高度，

然后就是裸记忆化暴搜了。。题解用的状态压缩来判断当前状态使用了几种高度以及这种高度以前有没有用过。。不是太懂状态压缩，然后我觉着开标记数组也可以解决这个问题。。然后交上去A掉了发现内存耗的好多。。

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                 using namespace std; ll dp[18][5][2][4]; bool vis[5]; int n; ll dfs(int x,int high,bool ok,int use) { if(dp[x][high][ok][use]!=-1) return dp[x][high][ok][use]; if(x==n) { if(use>=3&&ok) return 1; else return 0; } ll ans=0; for(int i=1;i<=4;i++) { if(!vis[i]) { vis[i]=1; if(ok||(high==1&&i==4)||(high==4&&i==1)) ans+=dfs(x+1,i,1,min(use+1,3)); else ans+=dfs(x+1,i,0,min(use+1,3)); vis[i]=0; } else { if(ok||(high==1&&i==4)||(high==4&&i==1)) ans+=dfs(x+1,i,1,min(use,3)); else ans+=dfs(x+1,i,0,min(use,3)); } } return dp[x][high][ok][use]=ans; } int main() { while(~scanf("%d",&n)&&n!=-1) { if(n<3) { printf("%d: 0\n",n); continue; } memset(dp,-1,sizeof(dp)); dfs(0,0,0,0); printf("%d: %lld\n",n,dp[0][0][0][0]); } return 0; }