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. 第3节 sqoop:5、实现数据的控制导入

    导入表数据子集 我们可以导入表的使用Sqoop导入工具,"where"子句的一个子集.它执行在各自的数据库服务器相应的SQL查询,并将结果存储在HDFS的目标目录. where子句 ...

  2. mongodb - 关联字段

    1,博客表结构  Blog.js var mongoose = require('mongoose') mongoose.connect('mongodb://localhost/test',{ us ...

  3. Pdo 绑定参数方法 通用更新语句

    public  static  function upShelf($table,$arr,$where)    { //将传入的数组进行key vlaue 分离 并将key值以问号方式绑定参数     ...

  4. Day4 - K - Ant Trip HDU - 3018

    Ant Country consist of N towns.There are M roads connecting the towns. Ant Tony,together with his fr ...

  5. Jupyter Notebooks usage

    Important note: You should always work on a duplicate of the course notebook. On the page you used t ...

  6. S7-300 实训3 异步电机正反转控制

    含有视频 方便以后查阅 参考书籍 跟我动手学 S7-300/400 PLC 第2版  廖常初 主编 实训3 异步电动机 正反转控制 步骤1 步骤2 在 cycle execution 前方 右击 插入 ...

  7. 小米手机收到升级鸿蒙OS提示?官方回应

    虽然尚未得到官方确认,但华为“鸿蒙”OS已经成为网络热门话题,在机圈引发热议. 本周,互联网上出现了显示为MIUI 10手机被锁定,屏幕上出现“小米将于2020年9月15日全面停止服务,届时您所有设备 ...

  8. 11.swoole学习笔记--进程信号触发器

    <?php //触发函数--异步执行 swoole_process::signal(SIGALRM,function(){ ; echo "$i \n"; $i++; ){ ...

  9. 125-PHP类__set()魔术方法

    <?php class ren{ //定义一个人类 //定义成员属性 private $name='tom'; private $age=15; //定义__set()魔术方法 public f ...

  10. 089-PHP数组运用 - 通过循环函数取出部分成员合并成新数组

    <?php function myfunc($arr){ //定义过滤函数 $j=count($arr); for($i=0;$i<$j;$i++){ if($arr[$i]>=0& ...