Flow Layout
Time Limit: 1000MS   Memory Limit: 30000K
Total Submissions: 3091   Accepted: 2148

Description

A flow layout manager takes rectangular objects and places them in a rectangular window from left to right. If there isn't enough room in one row for an object, it is placed completely below all the objects in the first row at the left edge, where the order
continues from left to right again. Given a set of rectangular dimensions and a maximum window width, you are to write a program that computes the dimensions of the final window after all the rectangles have been placed in it. 



For example, given a window that can be at most 35 units wide, and three rectangles with dimensions 10 x 5, 20 x 12, and 8 x 13, the flow layout manager would create a window that looked like the figures below after each rectangle was added. 




The final dimensions of the resulting window are 30 x 25, since the width of the first row is 10+20 = 30 and the combined height of the first and second rows is 12+13 = 25.

Input

The input consists of one or more sets of data, followed by a final line containing only the value 0. Each data set starts with a line containing an integer, m, 1 <= m <= 80, which is the maximum width of the resulting window. This is followed by at least one
and at most 15 lines, each containing the dimensions of one rectangle, width first, then height. The end of the list of rectangles is signaled by the pair -1 -1, which is not counted as the dimensions of an actual rectangle. Each rectangle is between 1 and
80 units wide (inclusive) and between 1 and 100 units high (inclusive).

Output

For each input set print the width of the resulting window, followed by a space, then the lowercase letter "x", followed by a space, then the height of the resulting window.

Sample Input

35
10 5
20 12
8 13
-1 -1
25
10 5
20 13
3 12
-1 -1
15
5 17
5 17
5 17
7 9
7 20
2 10
-1 -1
0

Sample Output

30 x 25
23 x 18
15 x 47

给了一个固定长度的空间,要往里面摆箱子,不够相应的空间箱子就要放到下一“行”,问最终的长度和宽度。宽度不一定是给定的宽度,可能有没有用掉的情况。

模拟水题。

代码:

#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <cstring>
#pragma warning(disable:4996)
using namespace std; int zong_weight,num;
int weight[20],height[20]; int main()
{
int i,sum,length,zong_length,max_weight;
while(scanf("%d",&zong_weight)==1)
{
if(zong_weight==0)
break; num=1;
sum=0;
length=0;
zong_length=0;
max_weight=0; while(scanf("%d%d",&weight[num],&height[num])==2)
{
if(weight[num]+height[num]==-2)
break;
num++;
} for(i=1;i<=num;i++)
{
if(sum+weight[i]<=zong_weight)
{
sum += weight[i];
length=max(length,height[i]);
max_weight=max(max_weight,sum);
}
else
{
sum=weight[i];
zong_length += length;
length=height[i];
}
}
zong_length += length;
cout<<max_weight<<" x "<<zong_length<<endl;
}
return 0;
}

版权声明:本文为博主原创文章,未经博主允许不得转载。

POJ 2014:Flow Layout 模拟水题的更多相关文章

  1. POJ 2014 Flow Layout 模拟

    http://poj.org/problem?id=2014 嘻嘻2014要到啦,于是去做Prob.ID 为2014的题~~~~祝大家新年快乐~~ 题目大意: 给你一个最大宽度的矩形,要求把小矩形排放 ...

  2. HDOJ 2317. Nasty Hacks 模拟水题

    Nasty Hacks Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Tota ...

  3. poj 3080 Blue Jeans(水题 暴搜)

    题目:http://poj.org/problem?id=3080 水题,暴搜 #include <iostream> #include<cstdio> #include< ...

  4. POJ 3984 - 迷宫问题 - [BFS水题]

    题目链接:http://poj.org/problem?id=3984 Description 定义一个二维数组: int maze[5][5] = { 0, 1, 0, 0, 0, 0, 1, 0, ...

  5. poj 1007:DNA Sorting(水题,字符串逆序数排序)

    DNA Sorting Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 80832   Accepted: 32533 Des ...

  6. poj 1004:Financial Management(水题,求平均数)

    Financial Management Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 126087   Accepted: ...

  7. POJ 3176 Cow Bowling (水题DP)

    题意:给定一个金字塔,第 i 行有 i 个数,从最上面走下来,只能相邻的层数,问你最大的和. 析:真是水题,学过DP的都会,就不说了. 代码如下: #include <cstdio> #i ...

  8. poj 1658 Eva's Problem(水题)

    一.Description Eva的家庭作业里有很多数列填空练习.填空练习的要求是:已知数列的前四项,填出第五项.因为已经知道这些数列只可能是等差或等比数列,她决定写一个程序来完成这些练习. Inpu ...

  9. hdu 5003 模拟水题 (2014鞍山网赛G题)

    你的一系列得分 先降序排列 再按0.95^(i-1)*ai 这个公式计算你的每一个得分 最后求和 Sample Input12530 478Sample Output984.1000000000 # ...

随机推荐

  1. git使用问题二删除远程仓库文件,本地保留不动

    git rm --cached filename/-r directory git commit "xxxx" git push

  2. 在GNOME开发人员的努力下,Pango 1.44即将问世

    早在5月份,Red Hat的Matthias Clasen共同制定了计划,在近年来相当陈旧的情况下,对Pango布局引擎库进行了一些改进. 这项工作将随着Pango1.44版本的发布而实现,看起来它很 ...

  3. day04-Python运维开发基础(位运算、代码块、流程控制)

    # (7)位运算符: & | ^ << >> ~ var1 = 19 var2 = 15 # & 按位与 res = var1 & var2 " ...

  4. 洛谷 P3435 [POI2006]OKR-Periods of Words

    题目传送门 解题思路: 这道题题面比较乱,先说一下这道题要求什么: 对于一个字符串,求它及它的所有前缀的一个答案串的长度之和,答案串就是对于一个字符串,找到一个它的一个前缀,这个前缀后面在复制一遍,得 ...

  5. mysql 添加索引语句

    1.PRIMARY  KEY(主键索引)        mysql>ALTER  TABLE  `table_name`  ADD  PRIMARY  KEY (  `column`  ) 2. ...

  6. 使用 json 模块,使json数据格式与Python字典dict数据格式互相转换,获取数据更加方便

    一.定义 JSON 是一种数据格式 使用 javaScript (Java 死鬼破特)对象表示法 二.特点 1.JSON 与 XML格式数据的区别 ====== 两种格式的数据,都是跨语言,跨平台 c ...

  7. NIO 聊天室代码实现

    服务器端 package com.ronnie.nio.groupChat; import java.io.IOException; import java.net.InetSocketAddress ...

  8. Python中eval与exec用法的区别

    Python中eval,exec这两个函数有着相似的输入参数类型和执行功能,因此在用法上经常出现混淆,以至经常用错,程序易抛出错误.下面主要通过这两个函数的语法来阐述区别,并用例子来进一步说明. 首先 ...

  9. Fr3设置图片打印

    见 fr3的文件内容,为xml <?xml version="1.0" encoding="utf-8"?> <TfrxReport Vers ...

  10. checkbox checked属性值

    记住我1<input type='checkbox' /> 记住我2<input type='checkbox' /> <button onclick='hehe();' ...