博客
关于我
AtCoder Beginner Contest 173 English C - H and V 二进制枚举
阅读量:146 次
发布时间:2019-02-28

本文共 1253 字,大约阅读时间需要 4 分钟。

题目链接:

因为我不会而且还是看了队友的代码才明白有二进制有这种所以发个文章记录一下错题

Problem Statement

We have a grid of H rows and W columns of squares. The color of the square at the i-th row from the top and the j-th column from the left (1≤i≤H,1≤j≤W) is given to you as a character ci,j: the square is white if ci, is., and black if ci,j is #.

Consider doing the following operation:

Choose some number of rows (possibly zero), and some number of columns (possibly zero). Then, paint red all squares in the chosen rows and all squares in the chosen columns.

You are given a positive integer K. How many choices of rows and columns result in exactly K black squares remaining after the operation? Here, we consider two choices different when there is a row or column chosen in only one of those choices.

Constraints

1≤H,W≤6

1≤K≤HW
ci,j is . or #.

题意:

在h*w的矩阵中含有‘.’和‘#’分别代表白色和黑色格子,现在你可以选择某一行或者某一列让格子变成红色,求有多少种可能使得染成红色后,还恰好剩下k个黑色格子。

题解:

for(int i=0;i<(1<<h);i++)

for(int j=0;j<(1<<w);j++)
二进制每次+1就可以暴力遍历每种情况出现的可能性

#include
using namespace std;#define ll long long#define endl "\n"int main(){ ios_base::sync_with_stdio(0);cin.tie(0); int n,m,k; cin>>n>>m>>k; int ans=0; bool ar[n][m]; for(int i=0;i
>x; ar[i][j]= x == '#'; } for(int i=0;i<(1<

转载地址:http://hdld.baihongyu.com/

你可能感兴趣的文章
MySQL-数据页的结构
查看>>
MySQL-架构篇
查看>>
MySQL-索引的分类(聚簇索引、二级索引、联合索引)
查看>>
Mysql-触发器及创建触发器失败原因
查看>>
MySQL-连接
查看>>
mysql-递归查询(二)
查看>>
MySQL5.1安装
查看>>
mysql5.5和5.6版本间的坑
查看>>
mysql5.5最简安装教程
查看>>
mysql5.6 TIME,DATETIME,TIMESTAMP
查看>>
mysql5.6.21重置数据库的root密码
查看>>
Mysql5.6主从复制-基于binlog
查看>>
MySQL5.6忘记root密码(win平台)
查看>>
MySQL5.6的Linux安装shell脚本之二进制安装(一)
查看>>
MySQL5.6的zip包安装教程
查看>>
mysql5.7 for windows_MySQL 5.7 for Windows 解压缩版配置安装
查看>>
Webpack 基本环境搭建
查看>>
mysql5.7 安装版 表不能输入汉字解决方案
查看>>
MySQL5.7.18主从复制搭建(一主一从)
查看>>
MySQL5.7.19-win64安装启动
查看>>