poj 2082 Terrible Sets (数据结构 ——栈 STL）
Terrible Sets
| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 2999 | Accepted: 1549 |
Description
Let N be the set of all natural numbers {0 , 1 , 2 , . . . }, and R be the set of all real numbers. wi, hi for i = 1 . . . n are some elements in N, and w0 = 0.
Define set B = {< x, y > | x, y ∈ R and there exists an index i > 0 such that 0 <= y <= hi ,∑ 0<=j<=i-1wj <= x <= ∑ 0<=j<=iwj}
Again, define set S = {A| A = WH for some W , H ∈ R + and there exists x0, y0 in N such that the set T = { < x , y > | x, y ∈ R and x0 <= x <= x0 +W and y0 <= y <= y0 + H} is contained in set B}.
Your mission now. What is Max(S)?
Wow, it looks like a terrible problem. Problems that appear to be terrible are sometimes actually easy.
But for this one, believe me, it's difficult.
Define set B = {< x, y > | x, y ∈ R and there exists an index i > 0 such that 0 <= y <= hi ,∑ 0<=j<=i-1wj <= x <= ∑ 0<=j<=iwj}
Again, define set S = {A| A = WH for some W , H ∈ R + and there exists x0, y0 in N such that the set T = { < x , y > | x, y ∈ R and x0 <= x <= x0 +W and y0 <= y <= y0 + H} is contained in set B}.
Your mission now. What is Max(S)?
Wow, it looks like a terrible problem. Problems that appear to be terrible are sometimes actually easy.
But for this one, believe me, it's difficult.
Input
The input consists of several test cases. For each case, n is given in a single line, and then followed by n lines, each containing wi and hi separated by a single space. The last line of the input is an single integer -1, indicating the end of input. You may assume that 1 <= n <= 50000 and w
1h
1+w
2h
2+...+w
nh
n
< 10
9.
Output
Simply output Max(S) in a single line for each case.
Sample Input
3 1 2 3 4 1 2 3 3 4 1 2 3 4 -1
Sample Output
12 14
Source
Shanghai 2004 Preliminary
最近几天出去玩去了，都没做题，也没发博客了，今天做题感觉也没什么状态了。。做题状态还是要保持好啊，所以今天就对一些基本的数据结构练习一些，对STL不怎么熟悉，就练习一些STL，找找状态,栈的基本操作还是要搞懂！！
参考《ACM/ICPC算法训练教程上的代码》，思路讲解可以看看http://www.cnblogs.com/hxsyl/archive/2012/08/16/2643015.html
题意就是给你一些紧贴着x轴的相互挨着的矩形，给定每个矩形的长宽，问他们可以形成的最大的矩形是多少？
大概的思路就是，用栈来做，如果矩形的高度是递增的，就让它入栈，如果即将入栈的高度小于栈顶元素的高度，就出栈，退到保持递增，在出栈的过程中统计到递增的可以达到的高度的最大面积，记录矩形的总长度。
每次读入一个矩形，若它比栈顶元素还高就直接进栈，否则不断将栈中元素弹栈，直到当前栈顶元素能够与读入的矩形满足高度递增。弹栈过程中累加弹出的元素的宽度，然后每弹出一个就判断当前弹出元素的高度x累加的宽度能否更新最大面积ans。然后以新的矩形作高，已经弹出栈的元素总宽度加上新矩形宽度作宽，把这个矩形插入到栈里。
最终栈肯定是一个单调的，只需要再把栈一个个弹空，弹栈过程中仍像上面那样计算即可。
下面是ac的代码：
#include
#include
#include
using namespace std;
struct rectangle //矩形的结构体
{
int h;
int w;
}data;
int main()
{
int n,ans,prior_h,total_w,area;
while(scanf("%d",&n)&&n!=-1)
{
ans=0;
stacks; //定义一个空栈
prior_h=0; //上次进栈的矩形高度
for(int i=0;iprior_h)
{
s.push(data);
}
else
{
total_w=0; //总宽度
area=0; //当前面积
while(!s.empty()&&s.top().h>data.h)
{
total_w+=s.top().w;
area=total_w*s.top().h;
if(area>ans)
ans=area;
s.pop();
}
total_w+=data.w;
data.w=total_w;
s.push(data); //新矩形进栈
}
prior_h=data.h;
}
total_w=0;
area=0;
while(!s.empty())
{
total_w+=s.top().w;
area=total_w*s.top().h;
if(area>ans)
ans=area;
s.pop();
}
printf("%d\n",ans);
}
return 0;
}