题意：一个 M x N 矩阵(M <= 1000, N <= 1000)，其中有些格子有障碍，求空白区的最大子矩阵的面积。
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1505
——>>枚举每一行，维护以该行为底的每一列的空白区的高，分别对每一行求解后取最大值。。
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];
注：输入数据不大正常，对于一个 F 或 R 字符，用 2 个 getchar(); 去读入会WA。。接着用长为 2 的字符数组去读入，会 RE。。用长为 3 的字符去读入，可以。。说明读入的数据中有连续的两个非空字符，并且FR数据并没有间隔一个位置或者行的最后有多余字符。。
另外，单调栈也恰恰是解决这一类问题的绝好结构。。（自己写的单调栈总算稳定成型了。。）
#include <cstdio>
#include <cstring>
#include <algorithm>using std::max;const int MAXN = 1000 + 10;int M, N;
int h[MAXN];
int L[MAXN], R[MAXN];void Dp()
{char s[3];int ret = 0;scanf("%d%d", &M, &N);
// getchar();memset(h, 0, sizeof(h));for (int row = 1; row <= M; ++row){for (int i = 1; i <= N; ++i){
// getchar();
// ch = getchar();scanf("%s", s);if (s[0] == 'F'){++h[i];}else{h[i] = 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, (R[i] - L[i] + 1) * h[i]);}}printf("%d\n", 3 * ret);
}int main()
{int K;scanf("%d", &K);while (K--){Dp();}return 0;
}
单调栈实现：
#include <cstdio>
#include <cstring>
#include <algorithm>using std::max;const int MAXN = 1000 + 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 M, N;
int h[MAXN];
int L[MAXN], R[MAXN];void Solve()
{char s[3];int ret = 0;scanf("%d%d", &M, &N);memset(h, 0, sizeof(h));for (int row = 1; row <= M; ++row){for (int i = 1; i <= N; ++i){scanf("%s", s);if (s[0] == 'F'){++h[i];}else{h[i] = 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, (R[i] - L[i] + 1) * h[i]);}}printf("%d\n", 3 * ret);
}int main()
{int K;scanf("%d", &K);while (K--){Solve();}return 0;
}