HDU-5077-NAND(DFS+打表)

Problem Description

Xiaoqiang entered the “shortest code” challenge organized by some self-claimed astrologists. He was given a boolean function taking n inputs (in C++):

bool f(bool x1, bool x2, bool x3){
//your code goes here
//return something
}

All possible inputs and expected outputs of this function have been revealed:



Xiaoqiang’s code must be like:

bool a = NAND(b, c);

where “a” is a newly defined variable,“b” and “c” can be a constant (0/1) or a function parameter (x1/x2/x3) or a previously defined variable. NAND is the “not-and” function:

NAND(b, c)=!(b&&c)

Because NAND is universal, Xiaoqiang knew that he could implement any boolean function he liked. Also, at the end of the code there should be a return statement:

return y;

where y can be a constant or a function parameter or a previously defined variable. After staring at the function for a while, Xiaoqiang came up with the answer:

bool a = NAND(x1, x2);
bool b = NAND(x2, x3);
bool y = NAND(a, b); return y;

Xiaoqiang wants to make sure that his solution is the shortest possible. Can you help him?

 

Input

The first line contains an integer T (T ≤ 20) denoting the number of the test cases.

For each test case, there is one line containing 8 characters encoding the truth table of the function.

 

Output

For each test case, output a single line containing the minimum number of lines Xiaoqiang has to write.

 

Sample Input

  

   1
00010011
  

 

Sample Output

  

   4


   

    

     Hint
    
Due to the small input domain, you can solve all the cases on your computer and submit a program with a table of all the answers.
   
    
  

 

Source

2014 Asia AnShan Regional Contest

 

思路：暴搜打表。 DFS的时候注意剪枝，递归的时候把i和j往下传，这是一个很有效的剪枝。

#include <stdio.h>
#define max(A,B)(A>B?A:B)

int num[1<<8],ans[1<<8],zh[256][256],cnt;
bool vis[1<<8];

void dfs(int dep,int x,int y)
{
    int i,j;

    if(ans[num[cnt-1]]>dep) ans[num[cnt-1]]=dep;

    if(dep==10)  return;

    for(i=x;i<cnt;i++)
    {
        for(j=(i==x?y:0);j<=i;j++)//这里是关键
        {
            if(!vis[zh[num[i]][num[j]]])
            {
                vis[zh[num[i]][num[j]]]=1;
                num[cnt++]=zh[num[i]][num[j]];

                dfs(dep+1,i,j);

                vis[zh[num[i]][num[j]]]=0;
                cnt--;
            }
        }
    }
}

int main()
{
    int i,j;

    num[0]=240;//00001111
    num[1]=204;//00110011
    num[2]=170;//01010101
    num[3]=0;//00000000
    num[4]=255;//11111111

    for(i=0;i<(1<<8);i++) ans[i]=11;

    vis[240]=vis[204]=vis[170]=vis[0]=vis[255]=1;
    ans[240]=ans[204]=ans[170]=ans[0]=ans[255]=1;

    for(i=0;i<256;i++) for(j=0;j<256;j++)
    {
        zh[i][j]=~(i&j);
        zh[i][j]&=0x000000FF;
    }

    cnt=5;

    dfs(1,0,0);

    for(i=0;i<(1<<8);i++)  printf("%d,",ans[i]);
}