POJ 2082(最大连续矩形面积)
Terrible Sets
| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 2428 | Accepted: 1215 |
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
测试用例的情况
解题分析：
//想骂人，简单的题意说的那么高深，特别是最后一句，估计管理员要阴笑啦
//大致题意：给定连续的矩形的宽和长，求出最大的连续矩形的面积
/*
维护一个栈中元素高度单调递增的栈，初始化栈中第一个元素高度宽度均为0，
然后每次读入一个矩形，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×
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。
*/
#include <iostream>
#include <cstring>
#include <stack>
using namespace std;
typedef struct Node
{
int w,h;
}Node;
int main()
{
int i,j,k,T;
stack <Node > s;
while(cin>>T,~T)
{
int max_area = 0;
int total_w,cur_area;
Node *rect = new Node[T+2];
for(i=0;i<T;i++)
{
cin>>rect[i].w>>rect[i].h;
if(s.empty())
s.push(rect[i]);
else
{
total_w=cur_area=0;
if(rect[i].h>=s.top().h)//此处是大于等于
s.push(rect[i]);
else
{
while(!s.empty())
{
if(rect[i].h<s.top().h)//此处只是小于
{
total_w += s.top().w;
if((cur_area=total_w*s.top().h)>max_area)
max_area = cur_area;
s.pop();
}
else
break;//跳出和继续下一次是不一样的
}
total_w += rect[i].w;
rect[i].w = total_w;
s.push(rect[i]);
}
}
}
total_w = cur_area = 0;
while(!s.empty())
{
total_w += s.top().w;
if((cur_area=total_w*s.top().h)>max_area)
max_area = cur_area;
s.pop();
}
cout<<max_area<<endl;
delete []rect;//加不加均AC
}
//system("pause");
s.clear();//没加也AC
return 0;
}