2014鞍山现场赛H题HDU5077（DFS减枝+打表）

NAND

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 65    Accepted Submission(s): 14

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

100010011

 

Sample Output

4

题意：RT

思路：这题简化题意就是，要求构造最少的NAND式子，使得输入x1,x2,x3，输出一个8位二进制数

            由于x1,x2,x3的所有组合满足0~8，那么可以将这三个数的8种值先按列压成3个8位二进制数（类似于搜索的时候开了8个栈，这样压以后只需一个栈，方便处理，减枝）

            x1,x2,x3的取值如下

            000

            001

            010

            011

            100

            101

            110

            111

            按列压成8位二进制，x1 : 00001111  x2 : 00110011  x3 : 01010101

            然后不难发现所有的NAND操作变成了~（a&b）

            搜索的时候将新值入栈，如果搜到重复的就直接跳过，这个用一个数组记录每个数是否存在就好了

            还有一个很重要的减枝是设置一个start变量，因为每次得到新的数是从当前栈里的元素两两进行NAND操作得到的

            而在DFS进入下一层的时候实际上队列中的有些元素已经两两运算过了，所以就不需要再算一次，start的含义是下一层DFS里的循环遍历应该从栈的哪个位置开始

            打完表花了15秒，感觉挺快的~