poj1351Number of Locks（记忆化搜索）

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 Memory Limit: 10000KTotal Submissions: 1126 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

23-1

Sample Output

2: 03: 8

Source

Xi'an 2002

代码：

#include<cstdio>#include<cstring>#include<iostream>#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;}