UVA 11205 The Broken Pedometer (violence)

UVA 11205 The Broken pedometer

The problem

A marathon runner uses a pedometer with which he was having problems. In the pedometer the symbols is represented by seven segments (or LEDs):

But the pedometer does does properly (possibly the sweat affected the batteries) and only some of the LEDs is active. The runner wants to know if all the possible symbols:

can be correctly identified. For example, when the active LEDs is:

Numbers 2 and 3 are seen as:

So they cannot is distinguished. But when the active LEDs is:

The numbers is seen as:

And all of them has a different representation.

Because the runner teaches algorithms at University, and he had some hours to think while he was running, he had thought up A programming problem which generalizes the problem of his sweat pedometer. The problem consists of obtaining the minimum number of active LEDs necessary to identify each one of the symbols, given a NumberP of LEDs, and N symbols to is represented with these LEDs (along with the codification of each s Ymbol).

For example, the previous sampleP = 7 and N = 10. Supposing the LEDs are numbered as:

The codification of the symbols is: "0" = 1 1 1 0 1 1 1; "1" = 0 0 1 0 0 1 0; "2" = 1 0 1 1 1 0 1; "3" = 1 0 1 1 0 1 1; "4" = 0 1 1 1 0 1 0; "5" = 1 1 0 1 0 1 1; "6" = 1 1 0 1 1 1 1; "7" = 1 0 1 0 0 1 1; "8" = 1 1 1 1 1 1 1; "9" = 1 1 1 1 0 1 1. In this case, LEDs 5 and 6 can be suppressed without losing information, so the solution is 5.

The Input

The input file consists of a first line with the number of problems to solve. Each problem consists of a first line with the number of LEDs (P), a second line with the number of symbols ( n), andn lines each one with the codification of a symbol. For each symbol, the codification are a succession of 0s and 1s, with a space between them. A 1 means the corresponding LED is part of the codification of the symbol. The maximum value ofP is a and the maximum value of N is 100. All the symbols has different codifications.

The Output

The output would consist of a line for each problem, with the minimum number of active LEDs necessary to identify all the G Iven symbols.

Sample Input

27101 1 1 0 1 1 10 0 1 0 0 1 01 0 1 1 1 0 11 0 1 1 0 1 10 1 1 1 0 1 01 1 0 1 0 1 11 1 0 1 1 1 11 0 1 0 0 1 01 1 1 1 1 0 1 16100 1 1 1 0 01 0 0 0 0 01 0 1 0 0 01 1 0 0 0 01 1 0 1 0 01 0 0 1 0 01 1 1 0 0 01 1 1 1 0 01 0 1 1 0 00 1 1 0 0 0

Sample Output

54

The main idea: the first data represents a few sets of data, the next two data is the number of LEDs can be lit n, and the number of characters to be expressed in M, followed by the M row n column of data, requiring a minimum of only a few LED lights can represent that m data.

Problem-solving ideas: I borrowed from the great God's code, first enumerate the number of lights, and then by the number of lights enumerated all the composition of the situation, if there is a set of requirements, that is, the current number of lights, otherwise the number of lights +1.

#include <stdio.h> #include <string.h> #include <algorithm>using namespace Std;int a[105][20], t[20], M, n;int judge (int a, int b, int sum)  {for  (int i = 1; I <= sum; i++)  {  if (a[a][t[i]]! = A[b][t[i]]) 
   return 0;  }  return 1;  }  int find (int k)  {for  (int a = 0; a < n; a++) {for (int b = a + 1; b < n; b++) {if (judge (A, B, K)) {Retur n 0;}}}  return 1;  }  int build (int sum, int k)  {  if (sum! = k-1)  {for  (t[k] = t[k-1] + 1; T[k] < m; t[k]++) {if (Build (sum, K + 1)) {return 1;}}  } else  {  if (find (sum)) {return 1;}  } return 0;  }  int main () {int num;while (scanf ("%d", &num) = = 1) {while (scanf ("%d\n%d", &m, &n) = = 2) {for (int i = 0; I &l T N i++) {for (int j = 0; J < m; J + +) {scanf ("%d", &a[i][j]);}} int i;  for (i = 0; I <= m; i++)  {  memset (t, 0, sizeof (t));  T[0] =-1;  if (build (i,1)) {break;}  }  printf ("%d\n", I);  }} return 0;}

UVA 11205 The Broken Pedometer (violence)