UVA 565 565 Pizza Anyone? （深搜 +位运算）

  Pizza Anyone? 

You are responsible for ordering a large pizza for you and your friends. Each of them has told you what he wants on a pizza and what he does not; of course they all understand that since there is only going to be one pizza, no one is likely to have all their requirements satisfied. Can you order a pizza that will satisfy at least one request from all your friends?

The pizza parlor you are calling offers the following pizza toppings; you can include or omit any of them in a pizza:

Input Code	Topping
A	Anchovies
B	Black Olives
C	Canadian Bacon
D	Diced Garlic
E	Extra Cheese
F	Fresh Broccoli
G	Green Peppers
H	Ham
I	Italian Sausage
J	Jalapeno Peppers
K	Kielbasa
L	Lean Ground Beef
M	Mushrooms
N	Nonfat Feta Cheese
O	Onions
P	Pepperoni

Your friends provide you with a line of text that describes their pizza preferences. For example, the line

+O-H+P;

reveals that someone will accept a pizza with onion, or without ham, or with pepperoni, and the line

-E-I-D+A+J;

indicates that someone else will accept a pizza that omits extra cheese, or Italian sausage, or diced garlic, or that includes anchovies or jalapenos.

Input

The input consists of a series of pizza constraints.

A pizza constraint is a list of 1 to 12 topping constraint lists each on a line by itself followed by a period on a line by itself.

A topping constraint list is a series of topping requests terminated by a single semicolon.

An topping request is a sign character (+/-) and then an uppercase letter from A to P.

Output

For each pizza constraint, provide a description of a pizza that satisfies it. A description is the string `` Toppings:  " in columns 1 through 10 and then a series of letters, in alphabetical order, listing the toppings on the pizza. So, a pizza with onion, anchovies, fresh broccoli and Canadian bacon would be described by:

Toppings: ACFO

If no combination toppings can be found which satisfies at least one request of every person, your program should print the string

No pizza can satisfy these requests.

on a line by itself starting in column 1.

Sample Input

+A+B+C+D-E-F-G-H;
-A-B+C+D-E-F+G+H;
-A+B-C+D-E+F-G+H;
.
+A+B+C+D;
+E+F+F+H;
+A+B-G;
+O+J-F;
+H+I+C;
+P;
+O+M+L;
+M-L+P;
.
+A+B+C+D;
+E+F+F+H;
+A+B-G;
+P-O;
+O+J-F;
+H+I+C;
+P;
+O;
+O+M+L;
-O-P;
+M-L+P;
.

Sample Output

Toppings:
Toppings: CELP
No pizza can satisfy these requests.

题意：一个披萨有16种配料。现在有几个人要定一份披萨。每个人都有一定需求，+代表要哪种配料，-代表不要。

要找出一种披萨。来满足所有人至少一个需求。。如果找不到就输出No pizza can satisfy these requests.

思路：一共16种配料。选与不选每种配料2种选择，一共就是2^16种情况。暴力枚举。。结果超时了。。 因为每种披萨都要进行判断。判断的过程算进去就超时了。。没想出比较好的方法。。看到别人用位运算。。自己也试了下。结果就过了。。不过跑了500多MS。。不知道那些0.00几ms的大神怎么做的。。

位运算：把每个人需要与不需要，存成一个16位2进制数。然后在从0枚举到2 ^16 - 1代表每种披萨。如果有满足条件

他们的或运算会>0。。（关于位运算。稍微看下就能明白的）。。。然后就枚举直到有一种披萨符合条件。把该二进制数转换成相应披萨的字母输出。如果没有。就输出No pizza can satisfy these requests.

#include <stdio.h>
#include <string.h>

struct Q
{
    int yes;
    int no;
} q[10005];

int num = 0;
int sta;
int out[20];
char str[105];

int main()
{
    while (gets(str) != NULL)
    {
	if (str[0] == '.')
	{
	    for (sta = 0; sta < (1 << 16); sta ++)
	    {
		int i;
		for (i = 0; i < num; i ++)
		{
		    if ((q[i].yes & sta) || (q[i].no & (~sta)))
			continue;
		    else
			break;
		}
		if (i == num)
		    break;
	    }
	    int nnum = 0;
	    for (int i = 0; i < 16; i ++)
	    {
		if (sta & (1 << i))
		    out[nnum ++] = i + 'A';
	    }
	    if (sta == (1 << 16))
		printf("No pizza can satisfy these requests.\n");
	    else
	    {
		printf("Toppings: ");
		for (int i = 0; i < nnum; i ++)
		    printf("%c", out[i]);
		printf("\n");
	    }
	    memset(q, 0, sizeof(q));
	    memset(out, 0, sizeof(out));
	    num = 0;
	    continue;
	}
	for (int i = 0; str[i] != ';'; i += 2)
	{
	    if (str[i] == '+')
		q[num].yes |= (1 << (str[i + 1] - 'A'));
	    if (str[i] == '-')
		q[num].no |= (1 << (str[i + 1] - 'A'));
	}
	num ++;
    }
    return 0;
}