Codeforces Round #576 (Div. 1) D. Rectangle Painting 1

D. Rectangle Painting 1

time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output

There is a square grid of size n×n. Some cells are colored in black, all others are colored in white. In one operation you can select some rectangle and color all its cells in white. It costs max(h,w) to color a rectangle of size h×w. You are to make all cells white for minimum total cost.

Input

The first line contains a single integer n (1≤n≤50) — the size of the square grid.
Each of the next n lines contains a string of length n, consisting of characters ‘.’ and ‘#’. The j-th character of the i-th line is ‘#’ if the cell with coordinates (i,j) is black, otherwise it is white.

Output

Print a single integer — the minimum total cost to paint all cells in white.

Examples

input

3
###
#.#
###

output

3

input

3
...
...
...

output

0

inputCopy

4
#...
....
....
#...

output

2

input

5
#...#
.#.#.
.....
.#...
#....

output

5

題意爲有一個方形的區域，有白塊和黑塊，一次可以將一個矩形區域內的所有方塊變成白色，代價爲max（m,n）,即矩形的長邊。求最小代價。
這個問題可以使用dp解決。主要就是將這個大問題分解成小問題。f[x][y][z][w]表示使（x,y）,(z,w)兩點之間的區域全變成白色所需要的最低代價。
這裏將整個問題不斷分解成兩個小問題，試探所有可能，來求區域最小。遞歸的方法比較節省代碼，同時更容易理解。

import java.io.BufferedInputStream;
import java.io.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.io.StreamTokenizer;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Scanner;
import java.util.Stack;
 
public class Main {
	static StreamTokenizer st = new StreamTokenizer(new BufferedInputStream(System.in));
	static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
	static PrintWriter pr = new PrintWriter(new BufferedOutputStream(System.out));
	static Scanner sc = new Scanner(System.in);
 
//		long tic = System.currentTimeMillis();
//		long toc = System.currentTimeMillis();
//		System.out.println("Elapsed time: " + (toc - tic) + " ms");
	static int f[][][][] = new int[55][55][55][55];
	static String s[];
 
	public static void main(String[] args) {
		int n = sc.nextInt();
		s = new String[n+3];
		for (int i = 1; i <= n; i++) {
			s[i] = " "+sc.next();
		}
		for (int i = 0; i < 55; i++) {
			for (int j = 0; j < 55; j++) {
				for (int k = 0; k < 55; k++) {
					for (int p = 0; p < 55; p++) {
						f[i][j][k][p] = -1;
					}
				}
			}
		}
		System.out.println(dfs(1, 1, n, n));
 
	}
 
	static int dfs(int x, int y, int z, int w) {
		if (f[x][y][z][w] != -1)
			return f[x][y][z][w];
		if (x == z && y == w)
			return f[x][y][z][w] = (s[x].charAt(y) == '#' ? 1 : 0);
		int ret = Math.max(z - x + 1, w - y + 1);
		for (int i = x; i < z; i++)
			ret = Math.min(dfs(x, y, i, w) + dfs(i + 1, y, z, w), ret);
		for (int i = y; i < w; i++)
			ret = Math.min(dfs(x, y, z, i) + dfs(x, i + 1, z, w), ret);
		return f[x][y][z][w] = ret;
	}
 
	static int nextNum() {
		try {
			st.nextToken();
		} catch (IOException e) {
			e.printStackTrace();
		}
		return (int) st.nval;
	}
 
	static int log2(long x) {
		int flag = 1;
		int log = 0;
		while (x > flag) {
			flag <<= 1;
			log++;
		}
		return log;
	}
 
	static long phi(int x) {
		long res = x;
		for (long i = 2; i <= Math.sqrt(x); i++) {
			if (x % i == 0) {
				while (x % i == 0)
					x /= i;
				res = res / i * (i - 1);
			}
		}
		if (x > 1)
			res = res / x * (x - 1);
		return res;
	}
 
	static long pow(long a, long b, long mod) {
		if (b == 0)
			return 1;
		if (b == 1)
			return a;
		if ((b & 1) == 0)
			return pow(a * a % mod, b >> 1, mod);
		else
			return a * pow(a * a % mod, b >> 1, mod) % mod;
	}
 
	static long gcd(long a, long b) {
		if (b == 0) {
			return a;
		} else {
			return gcd(b, a % b);
		}
	}
 
	static int nextInt() {
		try {
			st.nextToken();
		} catch (IOException e) {
			e.printStackTrace();
		}
		return (int) st.nval;
	}
 
	static double nextDouble() {
		try {
			st.nextToken();
		} catch (IOException e) {
			e.printStackTrace();
		}
		return st.nval;
	}
 
	static String next() {
		try {
			st.nextToken();
		} catch (IOException e) {
			e.printStackTrace();
		}
		return st.sval;
	}
 
	static long nextLong() {
		try {
			st.nextToken();
		} catch (IOException e) {
			e.printStackTrace();
		}
		return (long) st.nval;
	}
}
 
class Node {
	int a;
	int b;
	int c;
 
	public Node() {
	}
 
	public Node(int a, int b, int c) {
		super();
		this.a = a;
		this.b = b;
		this.c = c;
	}
 
	@Override
	public String toString() {
		return "Node [a=" + a + ", b=" + b + "]";
	}
}