洛谷P4147 玉蟾宫(悬线法dp)

P4147 玉蟾宫

题解：悬线$dp$，先预处理出悬线的左右边界。$h$数组记录能到的最高高度。然后$max(h(i,j)∗(Right(i,j)−Left(i,j)+1)$即可。

代码

#include<bits/stdc++.h>
const int N = 1010;
using namespace std;
int n,m,rectangle,a[N][N],Left[N][N],Right[N][N],h[N][N];

int main()
{
#ifndef ONLINE_JUDGE
    freopen("input.in","r",stdin);
#endif
    char c;
    cin>>n>>m;
    for(int i = 1; i <= n; ++i){
        for(int j = 1; j <= m; ++j){
            cin>>c;
            a[i][j] = c == 'F' ? 1 : 0;
            Left[i][j] = Right[i][j] = j;
            h[i][j] = 1;
        }
    }
    for(int i = 1; i <= n; ++i){
        for(int j = 2; j <= m; ++j){
            if(a[i][j] == a[i][j - 1] && a[i][j] == 1) Left[i][j] = Left[i][j - 1];
        }
    }
    for(int i = 1; i <= n; ++i){
        for(int j = m - 1; j >= 1; --j){
            if(a[i][j] == a[i][j + 1] && a[i][j] == 1) Right[i][j] = Right[i][j + 1];
        }
    }
    for(int i = 1; i <= m; ++i) Left[0][i] = 0, Right[0][i] = m + 1;
    for(int i = 1; i <= n; ++i)
        for(int j = 1; j <= m; ++j) {
            if(a[i][j] == a[i - 1][j] && i != 1 && a[i][j] == 1) {
                h[i][j] += h[i - 1][j];
                Left[i][j] = max(Left[i][j],Left[i - 1][j]);
                Right[i][j] = min(Right[i][j],Right[i - 1][j]);
            }
            rectangle = max(rectangle, h[i][j] * (Right[i][j] - Left[i][j] + 1));
        }
    cout<<rectangle * 3<<endl;
    return 0;
}