poj2559

15 9月 2015

Largest Rectangle in a Histogram

A histogram is a polygon composed of a sequence of rectangles aligned at a common base line. The rectangles have equal widths but may have different heights. For example, the figure on the left shows the histogram that consists of rectangles with the heights 2, 1, 4, 5, 1, 3, 3, measured in units where 1 is the width of the rectangles: Usually, histograms are used to represent discrete distributions, e.g., the frequencies of characters in texts. Note that the order of the rectangles, i.e., their heights, is important. Calculate the area of the largest rectangle in a histogram that is aligned at the common base line, too. The figure on the right shows the largest aligned rectangle for the depicted histogram.

Input

The input contains several test cases. Each test case describes a histogram and starts with an integer n, denoting the number of rectangles it is composed of. You may assume that1<=n<=100000. Then follow n integers h₁,...,h_n, where 0<=h_i<=1000000000. These numbers denote the heights of the rectangles of the histogram in left-to-right order. The width of each rectangle is 1. A zero follows the input for the last test case.

Output

For each test case output on a single line the area of the largest rectangle in the specified histogram. Remember that this rectangle must be aligned at the common base line.

Sample Input

7 2 1 4 5 1 3 3

4 1000 1000 1000 1000

Sample Output

4000

Hint

Huge input, scanf is recommended.

Source

Ulm Local 2003

题目类型：单调队列

算法分析：一道经典的使用单调队列思想解的题。坐标从小到大的分析，如果存在一个高度不超过先前的高度，则应该将先前高度的矩形块的面积出栈并计算，更新最大值。否则就一直入栈。最后将还在栈中的矩形块的面积计算出来并更新最大值

#include <set>
#include <bitset>
#include <list>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <string>
#include <vector>
#include <ios>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <algorithm>
#include <utility>
#include <complex>
#include <numeric>
#include <functional>
#include <cmath>
#include <ctime>
#include <climits>
#include <cstdarg>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cassert>
using namespace std;

const int INF = 0x7FFFFFFF;
const double EPS = 1e-10;
const double PI = 2 * acos (0.0);
const int MOD = 10000007;

const int maxn = 100000 + 66;


struct Node
{
    long long h, w;
    Node (long long hh, long long ww) : h (hh), w (ww) {}
    Node () {h = 0, w = 0;}
};

int main()
{
//    freopen ("aaa.txt", "r", stdin);

    int n;
    while (scanf ("%d", &n) != EOF)
    {
        if (n == 0)
            break;

        deque <Node> deq;

        long long maxval = -INF, cnt;
        int i, input;

        for (i = 0; i < n; i++)
        {
            scanf ("%d", &input);

            cnt = 0;
            while (!deq.empty() && deq.back().h >= input)
            {
                Node temp = deq.back ();
                deq.pop_back();
                if (maxval < temp.h * (temp.w + cnt))
                        maxval = temp.h * (temp.w + cnt);

                cnt += temp.w;
            }

            deq.push_back (Node (input, 1 + cnt));
        }

        cnt = 0;
        while (!deq.empty ())
        {
            Node temp = deq.back ();
            deq.pop_back ();
            if (maxval < temp.h * (temp.w + cnt))
                maxval = temp.h * (temp.w + cnt);

            cnt += temp.w;
        }

        printf ("%lld\n", maxval);
    }

return 0;
}

#include <set>

#include <bitset>

#include <list>

#include <map>

#include <stack>

#include <queue>

#include <deque>

#include <string>

#include <vector>

#include <ios>

#include <iostream>

#include <fstream>

#include <sstream>

#include <iomanip>

#include <algorithm>

#include <utility>

#include <complex>

#include <numeric>

#include <functional>

#include <cmath>

#include <ctime>

#include <climits>

#include <cstdarg>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cctype>

#include <cassert>

using namespace std;

const int INF = 0x7FFFFFFF;

const double EPS = 1e-10;

const double PI = 2 * acos (0.0);

const int MOD = 10000007;

const int maxn = 100000 + 66;

struct Node

{

long long h, w;

Node (long long hh, long long ww) : h (hh), w (ww) {}

Node () {h = 0, w = 0;}

};

int main()

{

// freopen ("aaa.txt", "r", stdin);

int n;

while (scanf ("%d", &n) != EOF)

{

if (n == 0)

break;

deque <Node> deq;

long long maxval = -INF, cnt;

int i, input;

for (i = 0; i < n; i++)

{

scanf ("%d", &input);

cnt = 0;

while (!deq.empty() && deq.back().h >= input)

{

Node temp = deq.back ();

deq.pop_back();

if (maxval < temp.h * (temp.w + cnt))

maxval = temp.h * (temp.w + cnt);

cnt += temp.w;

}

deq.push_back (Node (input, 1 + cnt));

}

cnt = 0;

while (!deq.empty ())

{

Node temp = deq.back ();

deq.pop_back ();

if (maxval < temp.h * (temp.w + cnt))

maxval = temp.h * (temp.w + cnt);

cnt += temp.w;

}

printf ("%lld\n", maxval);

}

return 0;

}

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