poj 2082 Terrible Sets

Terrible Sets

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 4188		Accepted: 2166

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 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

这道的题意真的很难懂啊；

我也是参见了大神的题解才懂得；

大概就像是底边在一条直线上且相邻的长方形；

问连续长方形的最大面积；

最大面积要么是单独的一个长方形；

要么是它与之前的长方形联合组成；

#include<iostream>
#include<algorithm>
#include<cstring>
#include<string>
#include<cstdio>
#include<stack>
using namespace std;
struct S
{
    int w;
    int h;
}s[50005];
stack<S> st;
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF&&n!=-1)
    {
        int maxs=0;
        scanf("%d %d",&s[0].w,&s[0].h);
        st.push(s[0]);
        for(int i=1;i<n;++i)
        {
            scanf("%d %d",&s[i].w,&s[i].h);
            if(s[i].h>st.top().h)
            {
                st.push(s[i]);
            }
            else
            {
                int temp=0;
                while(!st.empty()&&s[i].h<=st.top().h)
                {
                    temp+=st.top().w;
                    maxs=max(maxs,temp*st.top().h);
                    st.pop();
                }
                S t;
                t.w=temp+=s[i].w;
                t.h=s[i].h;
                st.push(t);
            }
        }
        int temp=0;
        while(!st.empty())
        {
            temp+=st.top().w;
            maxs=max(maxs,temp*st.top().h);
            st.pop();
        }
        printf("%d\n",maxs);
    }
    return 0;
}