杭电Acm-1045解题心得

这个题目比较好理解。在一个n*n的格子中，放尽可能多的碉堡。这个题目开始想的时候，如果就思考深度搜索或者宽度搜索，会得到错误的答案，因为，当两个碉堡之间有墙的时候，它们是可以在同一行，同一列出现的。也就是说，在遍历的过程中，一行一列是否可以放碉堡的状态是可变的。代码如下:

import java.util.Scanner;

public class P1045 {

	private static int n, max;
	private static byte[][] city;
	private static byte[][] used;
	private static byte[] hori, vert;
	private static void trans(int num) {
		for (int i = 0; i < n; ++i) {
			for (int j = 0; j < n; ++j) {
				if (0 == used[i][j] && 0 == hori[i] && 0 == vert[j]) {
					if (1 == used[i][j]) continue;
					used[i][j] = 1;
					hori[i] = 1;
					vert[j] = 1;
					trans(num + 1);
					used[i][j] = 0;
					hori[i] = 0;
					vert[j] = 0;
				} else if (1 == city[i][j]) {
					hori[i] = 0;
					vert[j] = 0;
				}
			}
		}
		if (num > max) max = num;
 	}
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		while (in.hasNextLine()) {
			n = Integer.parseInt(in.nextLine());
			if (0 == n) return;
			city = new byte[n][n];
			used = new byte[n][n];
			hori = new byte[n];
			vert = new byte[n];
			max = 0;
			for (int i = 0; i < n; ++i) {
				String line = in.nextLine();
				for (int j = 0; j < n; ++j) {
					char c = line.charAt(j);
					if ('X' == c) {
						used[i][j] = 1;
						city[i][j] = 1;
					}
				}
			}
			for (int i = 0; i < n; ++i) {
				for (int j = 0; j < n; ++j) {
					if (1 == used[i][j]) continue;
					used[i][j] = 1;
					hori[i] = 1;
					vert[j] = 1;
					trans(1);
					used[i][j] = 0;
					hori[i] = 0;
					vert[j] = 0;
				}
			}
			System.out.println(max);
		}
	}
}