[洛谷P4147] 玉蟾宫

类型：单调栈

传送门：>Here<

题意：求一个$01$矩阵中最大子矩形（全是$1$）的面积

解题思路

单调栈的一个经典应用

考虑维护一个数组$p[i][j]$表示$(i,j)$往上最多有多少个连续的$1$。于是问题就转化为上一题的问题了，$p$即为高度，往左右扩散，利用单调栈求即可。总复杂度$O(n^2)$

个人认为，会做这类题的关键还是彻底弄懂上一题

Code

/*By DennyQi 2018.8.17*/
#include <cstdio>
#include <queue>
#include <cstring>
#include <algorithm>
#define  r  read()
#define  Max(a,b)  (((a)>(b)) ? (a) : (b))
#define  Min(a,b)  (((a)<(b)) ? (a) : (b))
using namespace std;
typedef long long ll;
const int MAXN = 10010;
const int MAXM = 27010;
const int INF = 1061109567;
inline int read(){
    int x = 0; int w = 1; register int c = getchar();
    while(c ^ '-' && (c < '0' || c > '9')) c = getchar();
    if(c == '-') w = -1, c = getchar();
    while(c >= '0' && c <= '9') x = (x<<3) + (x<<1) + c - '0', c = getchar();return x * w;
}
int N,M,top,ans,cnt; char c[2];
int a[1010][1010],p[1010][1010],h[1010],w[1010];
int main(){
    scanf("%d %d",&N,&M);
    getchar();
    for(int i = 1; i <= N; ++i){
        for(int j = 1; j <= M; ++j){
            scanf("%s",c);
            a[i][j] = (c[0] == 'F' ? 1 : 0);
        }
    }
    for(int i = 1; i <= N; ++i){
        for(int j = 1; j <= M; ++j){
            if(a[i][j]){
                p[i][j] = p[i-1][j] + 1;
            }
            else{
                p[i][j] = 0;
            }
        }
    }
    for(int i = 1; i <= N; ++i){
        top = 0;
        for(int j = 1; j <= M; ++j){
            if(!top || p[i][j] > h[top]){
                h[++top] = p[i][j];
                w[top] = 1;
            }
            else{
                cnt = 0;
                while(top > 0 && p[i][j] <= h[top]){
                    cnt += w[top];
                    ans = Max(ans, h[top] * cnt);
                    --top;
                }
                h[++top] = p[i][j];
                w[top] = cnt+1;
            }
        }
        cnt = 0;
        while(top > 0){
            cnt += w[top];
            ans = Max(ans, h[top] * cnt);
            --top;
        }
    }
    printf("%d", ans * 3);
    return 0;
}