HDU 5319 Painter(枚举)
Painter
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 745 Accepted Submission(s): 345
so there are two kinds of draws, one is like ‘\’ , the other is like ‘/’. In each draw he choose arbitrary number of grids to draw. He always drew the first kind in red color, and drew the other kind in blue color, when a grid is drew by both red and blue,
it becomes green. A grid will never be drew by the same color more than one time. Now give you the ultimate state of the board, can you calculate the minimum time of draws to reach this state.
Each test case begins with an integer number n describe the number of rows of the drawing board.
Then n lines of string consist of ‘R’ ‘B’ ‘G’ and ‘.’ of the same length. ‘.’ means the grid has not been drawn.
1<=n<=50
The number of column of the rectangle is also less than 50.
Output
Output an integer as described in the problem description.
2
4
RR.B
.RG.
.BRR
B..R
4
RRBB
RGGB
BGGR
BBRR
3
6
pid=5324" target="_blank" style="color:rgb(26,92,200); text-decoration:none">5324
5322 5321 5320#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<stdlib.h> using namespace std; char pp[91][91];
int v[91][91];
int n,m; int main()
{
int T;
scanf("%d",&T);
while(T--)
{
int ans = 0;
int k;
memset(v,0,sizeof(v));
scanf("%d",&n);
for(int i=0;i<n;i++)
{
scanf("%s",pp[i]);
k = strlen(pp[i]); }
for(int i=0;i<n;i++)
{
for(int j=0;j<k;j++)
{
if((pp[i][j] == 'R' || pp[i][j] == 'G')&& v[i][j] == 0)
{
int ii = i,jj = j;
while((pp[ii][jj] == 'R' || pp[ii][jj] == 'G') && v[ii][jj] == 0)
{
v[ii][jj]++;
//printf("v[%d][%d] = %d\n",ii,jj,v[ii][jj]);
ii = ii + 1;
jj = jj + 1;
if(ii>=n || jj>=k)
{
break;
}
}
ans++;
}
}
}
//printf("ans = %d\n",ans);
for(int i=0;i<n;i++)
{
for(int j=0;j<k;j++)
{
if((pp[i][j] == 'B'&&v[i][j] == 0) || (pp[i][j] == 'G' && v[i][j] == 1))
{
int x = i;
int y = j;
//printf("x = %d y = %d\n",x,y);
while((pp[x][y] == 'B' && v[x][y] == 0) || (pp[x][y] == 'G' && v[x][y] == 1))
{
v[x][y]++;
//printf("v[%d][%d] = %d\n",x,y,v[x][y]);
x = x + 1;
y = y - 1;
if(x>=n || y<0)
{
break;
}
}
ans++;
}
}
}
printf("%d\n",ans);
}
return 0;
}
HDU 5319 Painter(枚举)的更多相关文章
- hdu 5319 Painter(杭电多校赛第三场)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5319 Painter Time Limit: 2000/1000 MS (Java/Others) ...
- HDU 5319 Painter (模拟)
题意: 一个画家画出一张,有3种颜色的笔,R.G.B.R看成'\',B看成'/',G看成这两种的重叠(即叉形).给的是一个矩阵,矩阵中只有4种符号,除了3种颜色还有'.',代表没有涂色.问最小耗费多少 ...
- HDU 5319 Painter
题意:红色从左上向右下涂,蓝色从右上向左下涂,既涂红色又涂蓝色就变成绿色,问最少涂几下能变成给的图. 解法:模拟一下就好了,注意细节. 代码: #include<stdio.h> #inc ...
- 模拟+思维 HDOJ 5319 Painter
题目传送门 /* 题意:刷墙,斜45度刷红色或蓝色,相交的成绿色,每次刷的是连续的一段,知道最终结果,问最少刷几次 模拟+思维:模拟能做,网上有更巧妙地做法,只要前一个不是一样的必然要刷一次,保证是最 ...
- hdu 2489(枚举 + 最小生成树)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2489 思路:由于N, M的范围比较少,直接枚举所有的可能情况,然后求MST判断即可. #include ...
- hdu 3118(二进制枚举)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3118 思路:题目要求是去掉最少的边使得图中不存在路径长度为奇数的环,这个问题等价于在图中去掉若干条边, ...
- hdoj 5319 Painter(模拟题)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5319 思路分析:假设颜色R表示为1,颜色B表示为2,颜色G表示为3,因为数据量较小,采用暴力解法即可, ...
- HDU 6351暴力枚举 6354计算几何
Beautiful Now Time Limit: 5000/2500 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)T ...
- HDU 5319
Painter Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Total Su ...
随机推荐
- widows&l ubuntu 简单互联——基于samba
只是简单访问对方文件,就不上SSH了 windows端访问ubuntu: 要共享的文件夹右键 → 本地网络共享:共享此目录. 会提示你安装一个协议包,即samba(但此处默认安装包不完整),安装成功后 ...
- CentOS 7 挂载ntfs磁盘格式的U盘
因为CentOS 默认不识别NTFS的磁盘格式,所以我们要借助另外一个软件来挂载,那就是ntfs-3g了 自带的yum源没有这个软件,要用第三方的软件源,这里我用的是阿里的epel. 1. 切换到系统 ...
- Vue beaforeCreate时获取data中的数据
异步获取即:通过 $this.$nextTick或者settimeout,这连dom都可以拿出来 beforeCreate() { this.$nextTick(function() { con ...
- Zend Studio 服务器根目录设置
在 Apache 服务器根目录里查找 \conf\httpd.conf 例如:C:\AppServ\Apache24\conf\httpd.conf 打开后查找 DocumentRoot 标记 修改调 ...
- Java 序列化Serializable详解(附详细例子)
Java 序列化Serializable详解(附详细例子) 1.什么是序列化和反序列化Serialization(序列化)是一种将对象以一连串的字节描述的过程:反序列化deserialization是 ...
- jquery.guide.js 新手指引
/*! * by xyb * 新版上线时候的黑色半透明镂空遮罩指引效果实现jQuery小插件 * 兼容到IE8+ * MIT使用协议,使用时候保留版权 * */ $.guide = function ...
- 洛谷——P3907 圈的异或
P3907 圈的异或 无向图$dfs$找环,并判断边权异或和是否为0 #include<iostream> #include<cstdio> #include<algor ...
- LAMP 服务器环境
学习PHP脚本编程语言之前,必须先搭建并熟悉开发环境,开发环境有很多种,例如LAMP.WAMP.MAMP等.这里我介绍一下LAMP环境的搭建,即Linux.Apache.MySQL.PHP环境. 一. ...
- ubuntu Android Studio以及SDK安装
先使用迅雷下载linux版的Android Studio以及SDK,下载地址是https://dl.google.com/dl/android/studio/ide-zips/1.1.0/androi ...
- mysql查询排名
student_work表 student_info表 sql语句:按grade从高到低排名 结果: