Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

The Problem

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

But the pedometer does not work properly (possibly the sweat affected the batteries) and only some of the LEDs are active. The runner wants to know if all the possible symbols:

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

numbers 2 and 3 are seen as:

so they cannot be distinguished. But when the active LEDs are:

the numbers are seen as:

and all of them have a different representation.

Because the runner teaches algorithms at University, and he has some hours to think while he is running, he has 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 number P of LEDs, and N symbols to be represented with these LEDs (along with the codification of each symbol).

For example, in the previous sample P = 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), and N lines each one with the codification of a symbol. For each symbol, the codification is 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 of P is 15 and the maximum value of N is 100. All the symbols have different codifications.

The Output

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

Sample Input

2
7
10
1 1 1 0 1 1 1
0 0 1 0 0 1 0
1 0 1 1 1 0 1
1 0 1 1 0 1 1
0 1 1 1 0 1 0
1 1 0 1 0 1 1
1 1 0 1 1 1 1
1 0 1 0 0 1 0
1 1 1 1 1 1 1
1 1 1 1 0 1 1
6
10
0 1 1 1 0 0
1 0 0 0 0 0
1 0 1 0 0 0
1 1 0 0 0 0
1 1 0 1 0 0
1 0 0 1 0 0
1 1 1 0 0 0
1 1 1 1 0 0
1 0 1 1 0 0
0 1 1 0 0 0

Sample Output

5
4

---

題目描述：
給不同的字符表示方式，藉由不同的 LED 燈顯示，LED 燈的個數可能會有所不同，即不一定是七段顯示器，接著有些燈可能會壞，而造成辨識模稜兩可。問最少用 LED 的個數不影響給定的字符辨識。

題目解法：
基本上窮舉所有可能 LED 可用可不用的情況，因此是 O(2^15) 為上限，接著在每個字符的 LED 表示法壓縮成二進制，那麼就可以直接使用 AND 運算。
如果進行 AND 運算都不重複的話，即可以做為一組解，然後解析有多少的 bit = 1，找最少個數即可。

#include <stdio.h>

int main() {
    int testcase;
    int n, m, i, j, k;
    int x, v;
    scanf("%d", &testcase);
    while(testcase--) {
        scanf("%d %d", &n, &m);
        int A[105];
        for(i = 0; i < m; i++) {
            v = 0;
            for(j = 0; j < n; j++) {
                scanf("%d", &x);
                v |= (1&x)<<j;
            }
            A[i] = v;
        }
        int ret = 0xffff, rr = 1<<n;
        int mark[32768] = {};
        for(i = 0; i < rr; i++) {
            int flag = 0;
            x = 0;
            for(j = 0; j < n; j++)
                x += (i>>j)&1;
            if(x >= ret)    continue;
            for(j = 0; j < m && !flag; j++) {
                v = A[j]&i;
                if(mark[v]) {
                    while(j >= 0) {
                        v = A[j]&i;
                        mark[v] = 0;
                        j--;
                    }
                    flag = 1;
                    break;
                }
                mark[v] = 1;
            }
            if(flag == 0) {
                x = 0;
                for(j = 0; j < n; j++)
                    x += (i>>j)&1;
                if(x < ret)
                    ret = x;
                for(j = 0; j < m; j++) {
                    v = A[j]&i;
                    mark[v] = 0;
                }
            }
        }
        printf("%d\n", ret);
    }
    return 0;
}