POJ-1601

Pizza Anyone?

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 514 Accepted: 181

Description

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. 

#include <stdio.h>#include <stdlib.h>#include <string.h>#define MAX 17char peoples[MAX][MAX];int  pizza[MAX], Toppings[MAX], min, peopleNum;int cmp(const void *a, const void *b){    return *(char *)a - *(char *)b;}int scmp(const void *a, const void *b){return strlen((char *)a) - strlen((char *) b);}int getPizza(void){    int c, operators, i, n;    n = 0;    while (1) {          i = 0;          while ((c = getchar()) != ';')                switch (c) {                   case '-' : operators = -1; break;                   case '+' : operators = 1; break;                   case '.' : peopleNum = n;getchar();return 1;                   case EOF : return 0;                   default : peoples[n][i++] = operators*(c-'A'+1); break;                }          qsort(peoples[n], i, sizeof(peoples[n][0]), cmp);          peoples[n][i] = 0;  // 消除回车输入  getchar();          n++;           }}void dfs(int n, int l){    int i, temp, temp2;        if (l > min)       return;    if (peopleNum == n && l < min) {       for (i = 0; i < MAX; i++)           Toppings[i] = pizza[i];   min = l;       return;    }        for (i = 0; peoples[n][i] != 0; i++) {        temp = abs(peoples[n][i]);        temp2 = pizza[temp];        if (!pizza[temp]) {           pizza[temp] = peoples[n][i];           if (peoples[n][i] > 0)              l++;        }        if (pizza[temp] == peoples[n][i]) {           dfs(n+1, l);           if (!temp2 && peoples[n][i] > 0)              l--;           pizza[temp] = temp2;        }    }}int main(){    int i;    while (getPizza()) {          min = 1000;          memset(pizza, 0, sizeof(pizza));          // 这个很重要， 没排列的，跟POJ上的答案就不一样，虽然结果也是最优的，          // 但你的最优跟POJ上的最优不一样，就WA，坑人啊   qsort(peoples, peopleNum, sizeof(peoples[0]), scmp);          dfs(0, 0);          if (min != 1000) {              printf("Toppings: ");              for (i = 0; i < MAX; i++)                  if (Toppings[i] > 0)                     printf("%c", Toppings[i]+'A'-1);          }else               printf("No pizza can satisfy these requests.");          printf("\n");  memset(peoples, 0, sizeof(peoples));    }    return 0;}