Largest Rectangle in a Histogram(dp)

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

题意：给出n个矩形的高度，每个矩形的宽都为1，求相邻的矩形能组合成的最大的矩形的面积。

思路：求出比第i个矩形大的最左边的矩形的位置 l[i], 及比第i个矩形大的最右边的矩形的位置 r[i], 则第i个矩形的面积 s = (r[i]-l[i]+1)*hign[i]。

        如果第i-1个矩形比第i个矩形大，则 l[i] 必定在 l[i-1]的 左边，同理，r[i]必定在 r[i+1]的右边。

#include <stdio.h>
#include <string.h>
#include <algorithm>
#define LL __int64
const int N=100002;
using namespace std;
LL high[N],l[N],r[N];
int main()
{
    int n;
    while(~scanf("%d",&n)&&n)
    {
        for (int i = 1; i <= n; i++)
            scanf("%I64d",&high[i]);
        for (int i = 1; i <= n; i++)
        {
            int pos = i;
            while(pos > 1 && high[pos-1] >= high[i])
            {
                pos = l[pos-1];//如果第i个矩形左边的矩形(pos-1)比i高，
                              //则pos移动到l[pos-1](即比第pos-1个矩形高的最左边的矩形的位置)
            }
            l[i] = pos;//l[i]表示比第i个矩形连续高的最左边的矩形位置
        }
        for (int i = n; i >= 1; --i)
        {
            int pos = i;
            while(pos < n && high[i] <= high[pos+1])
            {
                pos = r[pos+1];
            }
            r[i] = pos;//表示比第i个矩形连续高的最右边的矩形位置
        }
        LL maxn = 0;
        for (int i = 1; i <= n; i++)
        {
            maxn = max((r[i]-l[i]+1)*high[i],maxn);
        }
        printf("%I64d
",maxn);
    }
    return 0;
}

同类型的题：

http://acm.hdu.edu.cn/showproblem.php?pid=1505

#include <stdio.h>
#include <algorithm>
#include <string.h>
#define LL __int64
const int N=100002;
using namespace std;
LL l[N],r[N],high[N];

LL max_area(int m)
{
    for (int i = 1; i <= m; i++)
    {
        if(i > 1 && high[i] <= high[i-1])
        {
            l[i] = l[i-1];
        }
        else
            l[i] = i;
    }
    for (int i = m; i >= 1; i--)
    {

        if(i < m && high[i] <= high[i+1])
        {
            r[i] = r[i+1];
        }
        else
            r[i] = i;
    }
    LL maxn = 0;
    for (int i = 1; i <= m; i++)
    {
        maxn = max((r[i]-l[i]+1)*high[i],maxn);
    }
    return maxn;
}
int main()
{
    int t;
    char s[23];
    scanf("%d",&t);
    while(t--)
    {
        int n,m;
        LL ans = 0;
        scanf("%d %d",&n,&m);
        memset(high,0,sizeof(high));
        for (int i = 1; i <= n; i++)
        {
            for (int j = 1; j <= m; j++)
            {
                scanf("%s",s);
                if (s[0]=='F')
                    high[j]++;
                else
                    high[j] = 0;
            }
            ans = max(max_area(m),ans);//第一行到第i行最大的面积
        }
        printf("%I64d
",ans*3);
    }
    return 0;
}