The broken pedometer-纯暴力枚举

The broken pedometer

Time Limit: 3000MS

Memory Limit: Unknown

64bit IO Format: %lld & %llu

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Description

 The Broken Pedometer 

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

此题目最重要的是看懂，他的大概意思是给你个P代表着有多少位,给你个N代表着有多少个数

终于要你求。每个数改变同样的位（题目是直接删掉同样的位。即关掉LED，让他不能执行）。最多改变多少，剩下的位依然能够表示独特性。即每个数都不同样，都能够识别出来

所以直接用DFS枚举全部的情况（中间剪枝一小部分）就可以。

/*
Author: 2486
Memory: 0 KB		Time: 33 MS
Language: C++ 4.8.2		Result: Accepted
VJ RunId: 4153206		Real RunId: 15826224
Public:		No Yes
*/
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int n,p,t,maps[105][105],Max,ind[20],ds[100+5];
bool vis[20];
bool C(int m) {
    int s;
    memset(vis,false,sizeof(vis));
    for(int i=0; i<m; i++) {
        vis[ind[i]]=true;
    }
    for(int i=0; i<n; i++) {
        s=0;
        for(int j=0; j<p; j++) {
            if(!vis[j]) {
                s=s*2+maps[i][j];
            }
        }
        ds[i]=s;
    }
    sort(ds,ds+n);
    for(int i=1; i<n; i++) {
        if(ds[i]==ds[i-1])return false;
    }
    return true;
}
void dfs(int col,int num) {
    if(!C(num))return;
    if(col>=p) {
        Max=max(Max,num);
        return ;
    }
    ind[num]=col;
    dfs(col+1,num+1);
    dfs(col+1,num);
}
int main() {
    scanf("%d",&t);
    while(t--) {
        scanf("%d%d",&p,&n);
        for(int i=0; i<n; i++) {
            for(int j=0; j<p; j++) {
                scanf("%d",&maps[i][j]);
            }
        }
        Max=0;
        dfs(0,0);
        printf("%d\n",p-Max);
    }
    return 0;
}

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