记忆化搜索——POJ 1351

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Number of Locks

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 1123		Accepted: 548

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

#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<map>
#include<queue>
#include<stack>
#include<vector>
#include<algorithm>
#include<cstring>
#include<string>
#include<iostream>
const int MAXN=1000+10;
using namespace std;
long long num[20][5][5][2];
//num[i][j][k][e];  i表示到达第几个slot,j表示有多少种slot，k表示到达的slot的编号是（1,2,3,4），e表示是否存在相连的1和4
int n;

long long dfs(int a,int b,int c,int d,int e)
{
	if(a==n){
		if(d && b>=3) return 1;
		else return 0;
	}
	if(num[a][b][c][d]!=-1) return num[a][b][c][d];
	long long ans=0;
	for(int i=1; i<=4; i++){
		int x=0;
		if(c==1 && i==4) x=1;
		else if(c==4 && i==1) x=1;
		int y;
		if((1<<(i-1))&e) y=b;
		else y=b+1;
		if(y>3) y=3;
		ans+=dfs(a+1, y, i, d||x, e|(1<<(i-1)));
	}
	num[a][b][c][d]=ans;
	return ans;
}

int main()
{
	//freopen("in.txt","r",stdin);
	while(cin>>n, n!=-1)
	{
		memset(num,-1,sizeof(num));
		long long res=dfs(0,0,0,0,0);
		cout<<n<<": "<<res<<endl;
	}
	return 0;
}