题意：一个V * U的矩阵，每个元素有一个高度Hxy，问长不超过100，且最高值与最低值的差不超过C的子矩阵的最大面积(1 <= U <= 700, 1 <= V <= 700, -30,000 <= Hxy <= 30,000, 0 <= C <= 10 )。

题目链接：http://poj.org/problem?id=1156

——>>枚举子矩阵的左右宽度（保证枚举宽度不超过100，同时记录所枚举左右区间的每行的最大最小值），再枚举子矩阵的上下宽度（用单调队列优化判C）。

#include <cstdio>
#include <algorithm>
#include <cstring>

using std::min;
using std::max;

const int MAXN = 700 + 1;

int H[MAXN][MAXN];
int arrRowMin[MAXN];
int arrRowMax[MAXN];

struct DQ
{
    int nFront;
    int nTail;
    int arr[MAXN];

    DQ():nFront(0), nTail(0){}

    void Init()
    {
        nFront = nTail = 0;
    }

    void PushBackMin(int nId)
    {
        while (nFront != nTail && arrRowMin[nId] <= arrRowMin[arr[nTail - 1]])
        {
            nTail--;
        }
        arr[nTail++] = nId;
    }

    void PushBackMax(int nId)
    {
        while (nFront != nTail && arrRowMax[nId] >= arrRowMax[arr[nTail - 1]])
        {
            nTail--;
        }
        arr[nTail++] = nId;
    }

    void PopFront()
    {
        nFront++;
    }

    int Front()
    {
        return arr[nFront];
    }

    bool Empty()
    {
        return nFront == nTail;
    }
};

int ReadInt()
{
    int nRet = 0;
    int nFlag = 1;
    char chIn;

    chIn = getchar();
    if (chIn == '-')
    {
        nFlag = -1;
    }
    else
    {
        nRet = nRet * 10 + chIn - '0';
    }
    while ((chIn = getchar()) && chIn >= '0' && chIn <= '9')
    {
        nRet = nRet * 10 + chIn - '0';
    }

    return nRet * nFlag;
}

int main()
{
    int U, V, C;

    while (scanf("%d%d%d", &U, &V, &C) == 3)
    {
        getchar();
        for (int i = 1; i <= V; ++i)
        {
            for (int j = 1; j <= U; ++j)
            {
                H[i][j] = ReadInt();
            }
        }

        DQ dqMin;
        DQ dqMax;
        int nRet = 0;

        for (int nLcol = 1; nLcol <= U; ++nLcol)
        {
            memset(arrRowMin, 0x3f, sizeof(arrRowMin));
            memset(arrRowMax, 0xc0, sizeof(arrRowMax));
            for (int nRcol = nLcol; nRcol <= U && nRcol - nLcol + 1 <= 100; ++nRcol)
            {
                for (int i = 1; i <= V; ++i)
                {
                    arrRowMin[i] = min(arrRowMin[i], H[i][nRcol]);
                    arrRowMax[i] = max(arrRowMax[i], H[i][nRcol]);
                }

                int nWidth = nRcol - nLcol + 1;
                dqMin.Init();
                dqMax.Init();
                for (int i = 1, j = 1; j <= V && nWidth * (V - i + 1) > nRet; ++j)
                {
                    dqMin.PushBackMin(j);
                    dqMax.PushBackMax(j);

                    while (i <= j && !dqMax.Empty() && !dqMin.Empty() && arrRowMax[dqMax.Front()] - arrRowMin[dqMin.Front()] > C)
                    {
                        i++;
                        while (!dqMax.Empty() && dqMax.Front() < i)
                        {
                            dqMax.PopFront();
                        }
                        while (!dqMin.Empty() && dqMin.Front() < i)
                        {
                            dqMin.PopFront();
                        }
                    }
                    nRet = max(nRet, nWidth * (j - i + 1));
                }
            }
        }

        printf("%d\n", nRet);
    }

    return 0;
}

poj - 1156 - A STRIP OF LAND（枚举 + 单调队列 + 输入开挂）