zoukankan      html  css  js  c++  java
  • UVA 11205 The broken pedometer(子集枚举)

    B - The broken pedometer
    Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu
    Appoint description: 

    Description

    Download as PDF
     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




    题意大概是问你最少要多少根二极管就能把全部的排列区分出来。。。

    #include<iostream>
    #include<algorithm>
    #include<stdio.h>
    #include<queue>
    #include<vector>
    #include<string.h>
    using namespace std;
    int mat[105][20];
    int main()
    {
        //freopen("in.txt","r",stdin);
        int T;
        scanf("%d",&T);
        int p,n;
        while(T--)
        {
            scanf("%d%d",&p,&n);
            for(int i=1;i<=n;i++)
            for(int j=1;j<=p;j++)
            scanf("%d",&mat[i-1][j-1]);
            int ans=500;
            int choose[20];
            for(int i=1;i<(1<<p);i++)
            {
                bool flag=true;
                memset(choose,0,sizeof(choose));
                int k=0;
                for(k=0;(1<<k)<=i;k++)
                    if((1<<k)&i)choose[k]=1;
                for(int i_=0;flag&&i_<n;i_++)
                for(int j=i_+1;flag&&j<n;j++){
                    bool t=true;///同样的
                    for(int l=0;t&&l<k;l++)if(choose[l])
                    {
                        if(mat[i_][l]!=mat[j][l])t=false;///没有同样的
                    }
                    if(t)flag=false;///有同样的失败
                }
                if(flag){
                    int temp=0;
                    for(int c=0;c<k;c++)if(choose[c])temp++;
                    ans=min(ans,temp);
                }
            }
            printf("%d
    ",ans);
        }
        return 0;
    }



  • 相关阅读:
    nmake不是内部或外部命令,也不是可运行的程序
    MinGW下载和安装教程
    Qt接单
    C++ web 框架
    原型链
    ssh: Could not resolve hostname的一种解决方案
    oracle客户端安装配置
    linux安装go
    golang 为什么结构体方法要写成指针传递
    traceback模块
  • 原文地址:https://www.cnblogs.com/mfrbuaa/p/5381464.html
Copyright © 2011-2022 走看看