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
【题意】有N个连续的矩形，它们的底边沿着Z轴，宽度为Wi，高为Hi,现在要求这些矩形所能组成的面积最大的连续矩形面积是多少。
【分析】这道题原本正常做法也可以做，但是现在数据量很大，正常做法加优化也要O(n2)，所以就要用数据结构的只是了，正好单调栈的操作原理与这个很像，所以就可以用单调栈来做，首先用栈保存矩形，如果高度递增则不断入栈，如果遇到当前输入的比栈顶高度小，则从栈顶开始不断出栈并且计算最大面积，直到栈顶高度小于当前输入高度则停止出栈，并把开始出栈矩形的宽度累加得到totalw,把totalw和当前输入的矩形宽度相加得到当前输入矩形的宽度，并入栈，这样栈中保存的永远都是高度递增的矩形，最后输入完了之后如果栈不为空，则依次出栈并计算最大面积。
用时125ms；
【代码】
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<stack>
using namespace std;
const int N=5e4+5;
typedef long long ll;
int totalw;
struct no
{
int w;
int h;
}node,ss;
stack<no> s;
int max(int x,int y)
{
return x>y?x:y;
}
int main()
{
int t;
while(~scanf("%d",&t)&&t!=-1)
{
int w,h;
int ans=0;
while(t--)
{
scanf("%d %d",&w,&h);
if(s.empty()){//栈空则直接加入
node.w=w;
node.h=h;
s.push(node);
}
else
{
totalw=0;
if(h>s.top().h)//高度保持递增，则直接加入栈顶
{
ss.w=w;
ss.h=h;
s.push(ss);
}
else
{
while(!s.empty()&&s.top().h>h)//栈非空且输入高度小于栈顶高度，出栈，totalw记录出栈矩形宽度和
{
totalw+=s.top().w;
ans=max(ans,totalw*s.top().h);//求矩形面积最大值
s.pop();
}
//现在栈顶矩形高度一定是小于等于输入矩形高度的
//要使栈依然保持递增状态
node.w = w + totalw ;
node.h= h;
s.push(node);
}
}
}
totalw = 0;
while(!s.empty())//栈非空，则再次计算最大面积
{
totalw += s.top().w;
ans = max (ans,totalw*s.top().h);
s.pop();
}
printf("%d\n",ans);
}
return 0;
}
