| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 3017 | Accepted: 1561 |
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.
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
w1h1+w2h2+...+wnhn <
109.
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
#include"iostream"
#include"stack"
#include"cstdio"
using namespace std;
struct abc
{
int w;
int h;
} data;
int main()
{
int lasth,n,i,ans,curarea,totalw;
while(cin>>n&&n!=-1)
{
stack<abc> s;
ans=0;
lasth=0;
for(i=0;i<n;i++)
{
cin>>data.w>>data.h;
if(data.h>=lasth)
{
lasth=data.h;
s.push(data);
}
else
{
totalw=0;
curarea=0;
while(!s.empty()&&s.top().h>=data.h)
{
totalw+=s.top().w;
curarea=totalw*s.top().h;
if(curarea>ans)
ans=curarea;
s.pop();
}
totalw+=data.w;
data.w=totalw;
s.push(data);
}
}
totalw=0;
curarea=0;
while(!s.empty())
{
totalw+=s.top().w;
curarea=totalw*s.top().h;
if(curarea>ans)
ans=curarea;
s.pop();
}
cout<<ans<<endl;
}
return 0;
}