学步园

题意：求一个长为 n(1 <= n <= 100000) 的直方图的最大子矩形的面积。

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1506

——>>暴力枚举每个位置的左边界和右边界，时间复杂度为O(n ^ 2)。。T_T。。

还是枚举每个位置的左边界和右边界，但用 dp 来优化。。时间复杂度大大降低。。

状态：L[i] 表示第 i 个位置的左边界。

状态转移方程：L[i] = L[L[i] - 1];

状态：R[i] 表示第 i 个位置的右边界。

状态转移方程：R[i] = R[R[i] + 1];

最近非常喜欢开输入挂。。

另外，这是单调栈的练手题。。也来了一发。。

dp实现：

#include <cstdio>
#include <algorithm>

using std::max;

const int MAXN = 100000 + 10;

int n;
int h[MAXN];
int L[MAXN], R[MAXN];

int ReadInt()
{
    int ret = 0;
    char ch;

    while ((ch = getchar()) && ch >= '0' && ch <= '9')
    {
        ret = ret * 10 + ch - '0';
    }

    return ret;
}

void Read()
{
    getchar();
    for (int i = 1; i <= n; ++i)
    {
        h[i] = ReadInt();
    }
}

void Dp()
{
    long long ret = 0;

    for (int i = 1; i <= n; ++i)
    {
        L[i] = i;
        while (L[i] - 1 >= 1 && h[L[i] - 1] >= h[i])
        {
            L[i] = L[L[i] - 1];
        }
    }
    for (int i = n; i >= 1; --i)
    {
        R[i] = i;
        while (R[i] + 1 <= n && h[R[i] + 1] >= h[i])
        {
            R[i] = R[R[i] + 1];
        }
    }

    for (int i = 1; i <= n; ++i)
    {
        ret = max(ret, (long long)(R[i] - L[i] + 1) * h[i]);
    }

    printf("%I64d\n", ret);
}

int main()
{
    while (scanf("%d", &n) == 1 && n)
    {
        Read();
        Dp();
    }

    return 0;
}

单调栈实现：

#include <cstdio>
#include <algorithm>

using std::max;

const int MAXN = 100000 + 10;

struct MS
{
    int st[MAXN];
    int top;

    MS(): top(0) {}

    void Init()
    {
        top = 0;
    }

    void PushMin(int* A, int i)
    {
        while (top != 0 && A[i] <= A[st[top - 1]])
        {
            --top;
        }
        st[top++] = i;
    }

    int Size()
    {
        return top;
    }

    int Second()
    {
        return st[top - 2];
    }
};

int n;
int h[MAXN];
int L[MAXN], R[MAXN];

int ReadInt()
{
    int ret = 0;
    char ch;

    while ((ch = getchar()) && ch >= '0' && ch <= '9')
    {
        ret = ret * 10 + ch - '0';
    }

    return ret;
}

void Read()
{
    getchar();
    for (int i = 1; i <= n; ++i)
    {
        h[i] = ReadInt();
    }
}

void Solve()
{
    long long ret = 0;
    MS ms;

    for (int i = 1; i <= n; ++i)
    {
        ms.PushMin(h, i);
        if (ms.Size() == 1)
        {
            L[i] = 1;
        }
        else
        {
            L[i] = ms.Second() + 1;
        }
    }

    ms.Init();
    for (int i = n; i >= 1; --i)
    {
        ms.PushMin(h, i);
        if (ms.Size() == 1)
        {
            R[i] = n;
        }
        else
        {
            R[i] = ms.Second() - 1;
        }
    }

    for (int i = 1; i <= n; ++i)
    {
        ret = max(ret, (long long)(R[i] - L[i] + 1) * h[i]);
    }

    printf("%I64d\n", ret);
}

int main()
{
    while (scanf("%d", &n) == 1 && n)
    {
        Read();
        Solve();
    }

    return 0;
}