P4147 玉蟾宮

P4147 玉蟾宮

給定一個 \(N * M\) 的矩陣
求最大的全爲 \(F\) 的子矩陣

Solution

懸線法
限制條件爲轉移來的和如今的都爲 \(F\)

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\n", ans * 3);
    }
int main(){
    init();
    solve();
    return 0;
    }