【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[5]; inline int getans(int *a) { int i,top = 0,ans = 0,tmp; static int s[MAXN],w[MAXN]; s[0] = a[m+1] = -1; for (i = 1; i <= m + 1; i++) { if (a[i] >= s[top]) { s[++top] = a[i]; w[top] = 1; } else { tmp = 0; while (a[i] < s[top]) { tmp += w[top]; ans = max(ans,tmp*s[top]); top--; } s[++top] = a[i]; w[top] = tmp + 1; } } return ans; } int main() { scanf("%d",&T); while (T--) { ans = 0; scanf("%d%d",&n,&m); for (i = 1; i <= n; i++) { for (j = 1; j <= m; j++) { scanf("%s",&tmp); f[i][j] = (tmp[0] == 'F') ? (f[i-1][j] + 1) : (0); } } for (i = 1; i <= n; i++) ans = max(ans,getans(f[i])); printf("%d ",ans*3); } return 0; }