uva 565 Pizza Anyone?(二进制+搜索）

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.

题目大意：就是给出一些字符串， + 代表需要， - 代表不需要，要你求一个最小的序列使得这个序列满足上述所有需求（一个需求只要满足其实中任意一个就行）。

解题思路:如果用普通的表示方法去做这道题的话，时间上绝对超时。可是仔细想一下，没样物品无非就是拿与不拿的关系，所以我们引入位位运算来简化判断的时间。

#include <stdio.h>
#include <string.h>
#include <math.h>
#define N 1000

int n, ok, con[2][N], bo[N];
int Max = (1 << 16) - 1;

int handle(char str[]){
    int len = strlen(str);
    for (int i = 0; i < len - 1; i++){
	if (str[i] == '+')
	    con[1][n] += 1 << (str[++i] - 'A');
	else
	    con[0][n] += 1 << (str[++i] - 'A');
    }
}

bool judge(int k){
    for (int i = 0; i < n; i++){
	if (((k^Max) & con[0][i]) || (k & con[1][i]))
	    continue;
	else
	    return false;
    }
    return true;
}

int main(){
    char str[N];
    memset(con, 0, sizeof(con));
    n = ok = 0;
    while (gets(str)){
	if (strcmp(str, ".") != 0){
	    handle(str);
	    n++;
	    continue;
	}

	for (int i = 0; i <= Max; i++){
	    if (judge(i)){
		ok = 1;
		n = i;
		break;
	    }
	}

	if (ok){
	    printf("Toppings: ");
	    for (int i = 0; i <= 15; i++){
		if (n & (1 << i))
		    printf("%c", 'A' + i); 
	    }
	    printf("\n");
	}
	else
	    printf("No pizza can satisfy these requests.\n");

	// Init;
	memset(con, 0, sizeof(con));
	n = ok = 0;
    }
    return 0;}