首页 > 代码库 > poj1351Number of Locks（记忆化搜索）

poj1351Number of Locks（记忆化搜索）

2024-07-18 07:08:51   223人阅读

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: 10000K
Total 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

2
3
-1

Sample Output

2: 0
3: 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;
}




联系
我们