POJ1964 City Game

城市毒瘤。

求最大子矩阵，我首先考虑行列分开，觉得正解码量有点大就没想写。然后考虑前缀和/DP，发现也不行，最后无奈分析了一波复杂度，开始写正解。中间抄了一下之前写的一题的代码，然后WA了...原因是扫第二遍的时候减号左右两边写反了，还有就是一些奇奇怪怪的小细节，重新写一遍就A了。

#include <cstdio>
const int N = 1010;

int G[N][N], upto[N][N], l[N], temp[N], Q[N], top;

inline void max(int &a, int b) {
    if(a < b) a = b;
    return;
}
inline void min(int &a, int b) {
    if(a > b) a = b;
    return;
}

inline int solve(int n) {
    //printf("temp: ");
    //for(int i = 1; i <= n; i++)
    for(int i = 1; i <= n + 1; i++) {
        Q[++top] = i;
        while(temp[Q[top - 1]] > temp[i]) {
            l[Q[top - 1]] = i - Q[top - 1] - 1;
            Q[top - 1] = Q[top--];
        }
    }
    top = 0;
    for(int i = n; i >= 0; i--) {
        Q[++top] = i;
        while(temp[Q[top - 1]] > temp[i]) {
            l[Q[top - 1]] += Q[top - 1] - i;
            Q[top - 1] = Q[top--];
        }
    }
    int ans = 0;
    for(int i = 1; i <= n; i++) {
        max(ans, l[i] * temp[i]);
    }
    return ans;
}

int main() {
    int T;
    scanf("%d", &T);
    while(T--) {
        int m, n;
        char c;
        scanf("%d%d", &n, &m);

        top = temp[n + 1] = 0;

        for(int i = 1; i <= n; i++) {
            for(int j = 1; j <= m; j++) {
                c = getchar();
                while(c != 'R' && c != 'F') {
                    c = getchar();
                }
                //putchar(c);
                G[i][j] = (c == 'R') ? 1 : 0;
            }
        }
        //printf("in over\n");
        for(int j = 1; j <= m; j++) {
            int t = n + 1;
            for(int i = n; i >= 1; i--) {
                if(G[i][j]) {
                    t = i;
                }
                upto[i][j] = t;
            }
        }
        /*
        for(int i = 1; i <= n; i++) {
            for(int j = 1; j <= m; j++) {
                printf("%d ", upto[i][j]);
            }
            printf("\n");
        }
        */
        int now, ans = 0;
        for(int i = 1; i <= n; i = i + now + 1) {
            now = n;
            for(int j = 1; j <= m; j++) {
                temp[j] = upto[i][j] - i;
                min(now, temp[j]);
            }
            /*
            printf("%d %d \n", i, now);
            for(int j = 1; j <= m; j++) {
                printf("%d ", temp[j]);
            }
            puts("");
            */
            int p = solve(m);
            //printf("ans = %d \n", p);
            max(ans, p);
        }
        printf("%d\n", ans * 3);
    }

    return 0;
}