[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 2018.8.18*/
#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 = ;
const int MAXM = ;
const int INF = ;
inline int read(){
    int x = ; int w = ; register int c = getchar();
    while(c ^ '-' && (c < '' || c > '')) c = getchar();
    if(c == '-') w = -, c = getchar();
    while(c >= '' && c <= '') x = (x<<) + (x<<) + c - '', c = getchar();return x * w;
}
int N,M,top,ans,cnt; char c[];
int a[][],p[][],h[],w[];
int main(){
    scanf("%d %d",&N,&M);
    getchar();
    for(int i = ; i <= N; ++i){
        for(int j = ; j <= M; ++j){
            scanf("%s",c);
            a[i][j] = (c[] == 'F' ?  : );
        }
    }
    for(int i = ; i <= N; ++i){
        for(int j = ; j <= M; ++j){
            if(a[i][j]){
                p[i][j] = p[i-][j] + ;
            }
            else{
                p[i][j] = ;
            }
        }
    }
    for(int i = ; i <= N; ++i){
        top = ;
        for(int j = ; j <= M; ++j){
            if(!top || p[i][j] > h[top]){
                h[++top] = p[i][j];
                w[top] = ;
            }
            else{
                cnt = ;
                while(top >  && p[i][j] <= h[top]){
                    cnt += w[top];
                    ans = Max(ans, h[top] * cnt);
                    --top;
                }
                h[++top] = p[i][j];
                w[top] = cnt+;
            }
        }
        cnt = ;
        while(top > ){
            cnt += w[top];
            ans = Max(ans, h[top] * cnt);
            --top;
        }
    }
    printf("%d", ans * );
    return ;
}