P4147 玉蟾宫

P4147 玉蟾宫

给定一个 (N * M) 的矩阵
求最大的全为 (F) 的子矩阵

Solution

悬线法
限制条件为转移来的和现在的都为 (F)

Code

#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
#include<algorithm>
#include<climits>
#define LL long long
#define REP(i, x, y) for(int i = (x);i <= (y);i++)
using namespace std;
int RD(){
    int out = 0,flag = 1;char c = getchar();
    while(c < '0' || c >'9'){if(c == '-')flag = -1;c = getchar();}
    while(c >= '0' && c <= '9'){out = out * 10 + c - '0';c = getchar();}
    return flag * out;
    }
const int maxn = 2019;
int lenx, leny;
int map[maxn][maxn];
int l[maxn][maxn], r[maxn][maxn];
int up[maxn][maxn];
int ans;
void init(){
	lenx = RD(), leny = RD();
	REP(i, 1, lenx)REP(j, 1, leny){
		char s[19];
		scanf("%s", s);
		if(s[0] == 'F'){
			map[i][j] = 1;
			l[i][j] = r[i][j] = j;
			up[i][j] = 1;
			}
		}
	REP(i, 1, lenx)REP(j, 2, leny){
		if(map[i][j] && map[i][j - 1])l[i][j] = l[i][j - 1];
		}
	REP(i, 1, lenx)for(int j = leny - 1;j >= 1;j--){
		if(map[i][j] && map[i][j + 1])r[i][j] = r[i][j + 1];
		}
	}
void solve(){
	REP(i, 1, lenx)REP(j, 1, leny){
		if(!map[i][j])continue;
		if(i > 1 && map[i - 1][j]){
			l[i][j] = max(l[i][j], l[i - 1][j]);
			r[i][j] = min(r[i][j], r[i - 1][j]);
			up[i][j] = up[i - 1][j] + 1;
			}
		int a = r[i][j] - l[i][j] + 1;
		ans = max(ans, a * up[i][j]);
		}
	printf("%d
", ans * 3);
	}
int main(){
	init();
	solve();
	return 0;
	}