AtCoder Regular Contest 080 D - Grid Coloring
地址:http://arc080.contest.atcoder.jp/tasks/arc080_b
题目:
D - Grid Coloring
Time limit : 2sec / Memory limit : 256MB
Score : 400 points
Problem Statement
We have a grid with H rows and W columns of squares. Snuke is painting these squares in colors 1, 2, …, N. Here, the following conditions should be satisfied:
- For each i (1≤i≤N), there are exactly ai squares painted in Color i. Here, a1+a2+…+aN=HW.
- For each i (1≤i≤N), the squares painted in Color i are 4-connected. That is, every square painted in Color i can be reached from every square painted in Color i by repeatedly traveling to a horizontally or vertically adjacent square painted in Color i.
Find a way to paint the squares so that the conditions are satisfied. It can be shown that a solution always exists.
Constraints
- 1≤H,W≤100
- 1≤N≤HW
- ai≥1
- a1+a2+…+aN=HW
Input
Input is given from Standard Input in the following format:
H W
N
a1 a2 … aN
Output
Print one way to paint the squares that satisfies the conditions. Output in the following format:
c11 … c1W
:
cH1 … cHW
Here, cij is the color of the square at the i-th row from the top and j-th column from the left.
思路:蛇形填数即可
#include <bits/stdc++.h> using namespace std; #define MP make_pair
#define PB push_back
typedef long long LL;
typedef pair<int,int> PII;
const double eps=1e-;
const double pi=acos(-1.0);
const int K=1e6+;
const int mod=1e9+; int r,c,n,x,y,ans[][]; int main(void)
{
scanf("%d%d%d",&r,&c,&n);
x=,y=;
for(int i=,cnt;i<=n;i++)
{
scanf("%d",&cnt);
while(cnt--)
{
ans[x][y]=i;
if(y==c&&x%==)
y=c,x++;
else if(y==&&x%==)
y=,x++;
else if(x&)
y++;
else
y--;
}
}
for(int i=;i<=r;i++)
for(int j=;j<=c;j++)
printf("%d%c",ans[i][j],j==c?'\n':' ');
return ;
}
AtCoder Regular Contest 080 D - Grid Coloring的更多相关文章
- AtCoder Regular Contest 080 [CDEF]
C - 4-adjacent Time limit : 2sec / Memory limit : 256MB Problem Statement We have a sequence of leng ...
- AtCoder Regular Contest 080
手贱去开了abc,这么无聊.直接arc啊 C - 4-adjacent Time limit : 2sec / Memory limit : 256MB Score : 400 points Prob ...
- AtCoder Regular Contest 080 E - Young Maids
地址:http://arc080.contest.atcoder.jp/tasks/arc080_c 题目: E - Young Maids Time limit : 2sec / Memory li ...
- AtCoder Regular Contest 080 C - 4-adjacent
地址:http://arc080.contest.atcoder.jp/tasks/arc080_a 题目: C - 4-adjacent Time limit : 2sec / Memory lim ...
- AtCoder Regular Contest 080 E:Young Maids
题目传送门:https://arc080.contest.atcoder.jp/tasks/arc080_c 题目翻译 给你一个\(n\)的排列\(p\),一个空序列\(q\),你每次可以从\(p\) ...
- AtCoder Regular Contest 061 DSnuke's Coloring
http://arc061.contest.atcoder.jp/tasks/arc061_b 题意: H行W列的矩阵中,然后挖了n个洞,输出j(0-9)行,对于第i行输出,有多少个3*3区域中有i个 ...
- AtCoder Regular Contest 080 (ARC080) E - Young Maids 线段树 堆
原文链接http://www.cnblogs.com/zhouzhendong/p/8934377.html 题目传送门 - ARC080 E - Young Maids 题意 给定一个长度为$n$的 ...
- 【递归】【线段树】【堆】AtCoder Regular Contest 080 E - Young Maids
给你一个1~n的排列p,n是偶数,每次从中任选一对相邻的数出来,插到排列q的开头,如此循环,问你所能得到的字典序最小的排列q. 我们先确定q开头的两个数q1,q2,q1一定是p的奇数位的最小的数,而q ...
- AtCoder Regular Contest 095E - Symmetric Grid
$n \leq 12,m \leq 12$,$n$行$m$列小写字母,现可做无数次操作:交换两行:交换两列.问是否有可能把他变成中心对称的. 没有去想分组枚举的复杂度QAQ 行和列的操作顺序是随意的. ...
随机推荐
- asp.net后台cs中的JSON格式变量在前台Js中调用方法(前后台示例代码)
//后台cs代码: using System; using System.Collections.Generic; using System.Linq; using System.Web; using ...
- VS2008 对话框编辑器“即时预览”
之前在VS2008中利用资源编辑器修改完对话框资源后,总是重新编译一下,然后Ctrl+F5运行来预览修改的效果,不断修改,不断编译,导致很费时,效率低下. 今天,发现了一个很好用的功能“Test Di ...
- iOS TabBar添加阴影
效果图如下所示: 直接上代码 //移除顶部线条 self.tabBar.backgroundImage = [UIImage new]; self.tabBar.shadowImage = [UIIm ...
- Spring_day04--HibernateTemplate介绍_整合其他方式_Spring分模块开发
HibernateTemplate介绍 1 HibernateTemplate对hibernate框架进行封装, 直接调用HibernateTemplate里面的方法实现功能 2 HibernateT ...
- MediaPlayer播放音频,也可以播放视频
使用MediaPlayer播放音频或者视频的最简单例子: JAVA代码部分: public class MediaPlayerStudy extends Activity { private Butt ...
- 从WebView跳到普通View
本文转载至 http://pingguohe.net/2011/06/25/webview_to_nativeview/ 做网络ios应用难免要用到UIWebViewController,直接嵌入一个 ...
- LAMP集群项目五 项目备份
1.打包到本地 2.推送到备份服务器 3.删除若干天前的备份 ip=`awk '/IPADDR/' /etc/sysconfig/network-scripts/ifcfg-eth0 |awk -F ...
- 前端性能优化-减少http请求,dns预解析,减少repaint和reflow
前端性能优化方法: 一 . 减少http请求 (1)通过合并图片,减少请求,俗称css sprites(css精灵)css sprites (2)lazyload懒加载,在需要的时候再加载 1.定义: ...
- Python--paramiko库:连接远程服务器操作文件
import paramikofrom loggingutils.mylogger import logger as log class SSHConnection(object): def __in ...
- C#全角半角转换输出解决方法
Microsoft.VisualBasic 命名空间 Strings 模块 StrConv 函数就具有大写/小写.全角/半角.中文简体/繁体等转换功能,字符串转换应该说是VB.NET的强项,是这样的: ...