UVA 11205 The broken pedometer(子集枚举)

简介:
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\n",ans);
    }
    return 0;
}






本文转自mfrbuaa博客园博客,原文链接:http://www.cnblogs.com/mfrbuaa/p/5381464.html,如需转载请自行联系原作者

相关文章
|
4天前
|
存储 关系型数据库 分布式数据库
PostgreSQL 18 发布,快来 PolarDB 尝鲜!
PostgreSQL 18 发布,PolarDB for PostgreSQL 全面兼容。新版本支持异步I/O、UUIDv7、虚拟生成列、逻辑复制增强及OAuth认证,显著提升性能与安全。PolarDB-PG 18 支持存算分离架构,融合海量弹性存储与极致计算性能,搭配丰富插件生态,为企业提供高效、稳定、灵活的云数据库解决方案,助力企业数字化转型如虎添翼!
|
15天前
|
弹性计算 关系型数据库 微服务
基于 Docker 与 Kubernetes(K3s)的微服务:阿里云生产环境扩容实践
在微服务架构中,如何实现“稳定扩容”与“成本可控”是企业面临的核心挑战。本文结合 Python FastAPI 微服务实战,详解如何基于阿里云基础设施,利用 Docker 封装服务、K3s 实现容器编排,构建生产级微服务架构。内容涵盖容器构建、集群部署、自动扩缩容、可观测性等关键环节,适配阿里云资源特性与服务生态,助力企业打造低成本、高可靠、易扩展的微服务解决方案。
1313 5
|
2天前
|
监控 JavaScript Java
基于大模型技术的反欺诈知识问答系统
随着互联网与金融科技发展,网络欺诈频发,构建高效反欺诈平台成为迫切需求。本文基于Java、Vue.js、Spring Boot与MySQL技术,设计实现集欺诈识别、宣传教育、用户互动于一体的反欺诈系统,提升公众防范意识,助力企业合规与用户权益保护。
|
14天前
|
机器学习/深度学习 人工智能 前端开发
通义DeepResearch全面开源!同步分享可落地的高阶Agent构建方法论
通义研究团队开源发布通义 DeepResearch —— 首个在性能上可与 OpenAI DeepResearch 相媲美、并在多项权威基准测试中取得领先表现的全开源 Web Agent。
1356 87
|
2天前
|
JavaScript Java 大数据
基于JavaWeb的销售管理系统设计系统
本系统基于Java、MySQL、Spring Boot与Vue.js技术,构建高效、可扩展的销售管理平台,实现客户、订单、数据可视化等全流程自动化管理,提升企业运营效率与决策能力。
|
4天前
|
弹性计算 安全 数据安全/隐私保护
2025年阿里云域名备案流程(新手图文详细流程)
本文图文详解阿里云账号注册、服务器租赁、域名购买及备案全流程,涵盖企业实名认证、信息模板创建、域名备案提交与管局审核等关键步骤,助您快速完成网站上线前的准备工作。
194 82
2025年阿里云域名备案流程(新手图文详细流程)