POJ-2081 Terrible Sets（暴力，单调栈）

Terrible Sets Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 4113 Accepted: 2122 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

题目说的乱七八糟看不懂，看别人的题解上面的题意才知道要干什么。这道题目有O(n)的算法就是运用单调栈，我写的暴力的方法枚举矩形，向左右扫描。

#include <iostream>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#include <algorithm>

using namespace std;
#define MAX 50000
struct Node
{
    int w;
    int h;
}a[MAX+5];
int n;
int ans;
int sum;
int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        if(n==-1)
            break;
        ans=0;
        sum=0;
        for(int i=0;i<n;i++)
            scanf("%d%d",&a[i].w,&a[i].h);
        for(int i=0;i<n;i++)
        {
            sum=0;
            for(int j=i+1;j<n;j++)
            {
                if(a[j].h>=a[i].h)
                    sum+=a[i].h*a[j].w;
                else
                    break;
            }
            for(int p=i-1;p>=0;p--)
            {
                if(a[p].h>=a[i].h)
                    sum+=a[i].h*a[p].w;
                else
                    break;
            }
            sum+=a[i].w*a[i].h;
            ans=max(ans,sum);

        }
        printf("%d\n",ans);
    }
    return 0;


}