UVa 11205 The broken pedometer (枚举好题&巧用二进制)

11205 - The broken pedometer

Time limit: 3.000 seconds

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=107&page=show_problem&problem=2146

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

巧用二进制判断两个数在某一种有效LED组(active LEDs)下是否显示是一样的。

都写在注释里了。

完整代码：

/*0.116s*/

#include <cstdio>
#include <cstring>
const int P = 1 << 15, maxn = 105;

bool vis[P + 5];
int a[maxn];

int count(int n)///统计n的二进制中1的个数
{
	int sum = 0;
	while (n)
	{
		if (n & 1)
			++sum;
		n >>= 1;
	}
	return sum;
}

int main()
{
	int t, p, n, x, i, j, ans, up;
	scanf("%d", &t);
	while (t--)
	{
		scanf("%d%d", &p, &n);
		for (i = 0; i < n; i++)
		{
			scanf("%d", &a[i]);
			for (j = 1; j < p; j++)
			{
				scanf("%d", &x);
				a[i] = (a[i] << 1) + x;///把01串压缩成一个数
			}
		}
		ans = maxn, up = 1 << p;
		for (i = 0; i < up; i++)///枚举
		{
			int bits = count(i);
			if (bits >= ans) continue;
			memset(vis, 0, sizeof(vis));
			for (j = 0; j < n; ++j)
			{
				int pos = a[j] & i;///a[j]在i下的显示结果
				if (vis[pos])///说明有两个数显示是一样的
					break;
				vis[pos] = true;
			}
			if (j < n) continue;
			ans = bits;
		}
		printf("%d\n", ans);
	}
	return 0;
}