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;
}