【POJ 1964】 City Game

【题目链接】

http://poj.org/problem?id=1964

【算法】

记f[i]表示第i行最多向上延伸的行数

然后，对于每一行，我们用单调栈计算出这一行向上延伸的最大矩形面积，取最大值，即可

【代码】

#include <algorithm>
#include <bitset>
#include <cctype>
#include <cerrno>
#include <clocale>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <deque>
#include <exception>
#include <fstream>
#include <functional>
#include <limits>
#include <list>
#include <map>
#include <iomanip>
#include <ios>
#include <iosfwd>
#include <iostream>
#include <istream>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <stdexcept>
#include <streambuf>
#include <string>
#include <utility>
#include <vector>
#include <cwchar>
#include <cwctype>
#include <stack>
#include <limits.h>
using namespace std;
#define MAXN 1010

int T,n,m,i,j,ans;
int f[MAXN][MAXN];
char tmp[];

inline int getans(int *a)
{
        int i,top = ,ans = ,tmp;
        static int s[MAXN],w[MAXN];
        s[] = a[m+] = -;
        for (i = ; i <= m + ; i++)
        {
                if (a[i] >= s[top])
                {
                        s[++top] = a[i];
                        w[top] = ;
                }    else
                {
                        tmp = ;
                        while (a[i] < s[top])
                        {
                                tmp += w[top];
                                ans = max(ans,tmp*s[top]);
                                top--;
                        }
                        s[++top] = a[i];
                        w[top] = tmp + ;
                }
        }
        return ans;
}

int main()
{

        scanf("%d",&T);
        while (T--)
        {
                ans = ;
                scanf("%d%d",&n,&m);
                for (i = ; i <= n; i++)
                {
                        for (j = ; j <= m; j++)
                        {
                                scanf("%s",&tmp);
                                f[i][j] = (tmp[] == 'F') ? (f[i-][j] + ) : ();
                        }
                }
                for (i = ; i <= n; i++) ans = max(ans,getans(f[i]));
                printf("%d\n",ans*);
        }

        return ;

}