POJ 2082 Terrible Sets 51nod 1102 面积最大的矩形 《题意好难懂---<贪心+单调栈>》

​​Terrible Sets​​

Time Limit: 1000MS

Memory Limit: 30000KB

64bit IO Format: %lld & %llu

​​Submit​​​ ​​Status​​

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​​

题意不好理解--题意理解了就简单多了-.-

n个矩形连在一起-问我们可以截的最大的矩形面积是多少-.-

附上学长的图文结合的样例分析--​​点击打开链接​​

代码：

#include<cstdio>
#include<stack>
#include<cstring>
#include<algorithm>
using namespace std;
struct node {
    int w,h;
}chu;
int main()
{
    int n;
    while (scanf("%d",&n),n!=-1)
    {
        int ws,hs,ww,hh,ss=0;
        stack<node > sta;
        chu.h=0;chu.w=0;
        sta.push(chu);
        while (n--)
        {
            scanf("%d%d",&ww,&hh);
            if (hh>sta.top().h)
            {
                chu.h=hh;chu.w=ww;
                sta.push(chu);
            }
            else
            {
                ws=0;
                while (hh<=sta.top().h)
                {
                    chu=sta.top();
                    sta.pop();
                    ws+=chu.w;
                    ss=max(ss,ws*chu.h);
                }
                ww+=ws;
                chu.h=hh;chu.w=ww;
                sta.push(chu);
            }
        }
        ws=0;
        while (!sta.empty())
        {
            chu=sta.top();
            sta.pop();
            ws+=chu.w;
            ss=max(ss,ws*chu.h);
        }
        printf("%d\n",ss);
    }
    return 0;
}

51nod 1102 面积最大的矩形链接：​​点击打开链接​​

代码和上面的差不多-.-稍微改一些-.-把输入ww--直接改成ww=1;并用longlong就行了-.-

代码：

#include<cstdio>
#include<stack>
#include<cstring>
#include<algorithm>
using namespace std;
#define LL long long
struct node {
    LL w,h;
}chu;
int main()
{
    int n;
    scanf("%d",&n);
    LL ws,hs,ww,hh,ss=0;
    stack<node > sta;
    chu.h=0;chu.w=0;
    sta.push(chu);
    while (n--)
    {
        scanf("%lld",&hh);
        ww=1;
        if (hh>sta.top().h)
        {
            chu.h=hh;chu.w=ww;
            sta.push(chu);
        }
        else
        {
            ws=0;
            while (hh<=sta.top().h)
            {
                chu=sta.top();
                sta.pop();
                ws+=chu.w;
                ss=max(ss,ws*chu.h);
            }
            ww+=ws;
            chu.h=hh;chu.w=ww;
            sta.push(chu);
        }
    }
    ws=0;
    while (!sta.empty())
    {
        chu=sta.top();
        sta.pop();
        ws+=chu.w;
        ss=max(ss,ws*chu.h);
    }
    printf("%lld\n",ss);
    return 0;
}