1506题意：给你连续的直方图（底边边长为1），求连续的矩阵面积。

对每个直方图，分别向左向右进行扩展。

#include<cstdio>
#include<stdlib.h>
#include<string.h>
#include<string>
#include<map>
#include<cmath>
#include<iostream>
#include <queue>
#include <stack>
#include<algorithm>
#include<set>
using namespace std;
#define INF 1e8
#define eps 1e-8
#define LL long long
#define N 100010
#define mol 1000000007
int i,n,t;  
LL a[N],l[N],r[N],Max;  
int main()  
{  
    while (scanf("%d",&n) &&  n)  
    {  
        for (i=1; i<=n; ++i) scanf("%I64d",&a[i]);  
        l[1]=1;  
        r[n]=n;  
        for(i=2;i<=n;i++)
        {
            t=i;
            while(t>1&&a[i]<=a[t-1])//从左往右向左扩展
                t=l[t-1];
            l[i]=t;
        }
        for(i=n-1;i>=1;i--)
        {
            t=i;
            while(t<n&&a[i]<=a[t+1])//从右往左向右扩展
                t=r[t+1];
            r[i]=t;
        }
        Max=0;  
        for (i=1; i<=n; ++i)  
        {  
            if ((r[i]-l[i]+1)*a[i]>Max) 
                Max=(r[i]-l[i]+1)*a[i];  
        }  
        printf("%I64d\n",Max);  
    }  
    return 0;  
}  

#include<cstdio>
#include<stdlib.h>
#include<string.h>
#include<string>
#include<map>
#include<cmath>
#include<iostream>
#include <queue>
#include <stack>
#include<algorithm>
#include<set>
using namespace std;
#define INF 1e8
#define eps 1e-8
#define LL long long
#define maxn 1001
#define mol 1000000007
int t,n,m,a[maxn][maxn],l[maxn][maxn],r[maxn][maxn];
char s[maxn];
int main()
{
	
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%d",&n,&m);
		int i,j;
		memset(a,0,sizeof(a));
		for(i=1;i<=n;i++)
		{
			for(j=1;j<=m;j++)
			{
				scanf("%s",s);
				if(s[0]=='F') a[i][j]=a[i-1][j]+1;
				else a[i][j]=0;
			}
		}
		//printf("\n");
		int ans=0;
		for(i=1;i<=n;i++)
		{
			int t;
			l[i][1]=1;r[i][m]=m;
			for(j=2;j<=m;j++)
			{
				t=j;
				while(t>1&&a[i][j]<=a[i][t-1])
					t=l[i][t-1];
				l[i][j]=t;
			}
			for(j=m-1;j>0;j--)
			{
				t=j;
				while(t<m&&a[i][j]<=a[i][t+1])
					t=r[i][t+1];
				r[i][j]=t;
			}
			for(j=1;j<=m;j++)
			{
				if(a[i][j]*(r[i][j]-l[i][j]+1)>ans) 
					ans=a[i][j]*(r[i][j]-l[i][j]+1);
			}
		}
		printf("%d\n",ans*3);
	}
	return 0;
}
/*
2
5 6
R F F F F F
F F F F F F
R R R F F F
F F F F F F
F F F F F F

5 5
R R R R R
R R R R R
R R R R R
R R R R R
R R R R R
*/