poj 2488 深搜+回溯

package com.liang.poj;

import java.util.Scanner;

public class Test2488 {

static boolean b = true;

public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
int x = scan.nextInt();
int y = scan.nextInt();

int[][] chess = new int[x][y];
boolean[][] visited = new boolean[x][y];
int[][] ways = new int[x][y];

if (x == 1 && y == 1) {
b = false;
System.out.println(0);
}

for (int i = 0; i < chess.length; i++) {
for (int j = 0; j < chess[i].length; j++) {
dfs(0, i, j, chess, visited, ways);
}
}

if (b) {
System.out.println("不存在这样的点");
}

}

public static void dfs(int level, int x, int y, int[][] chess,
boolean[][] visited, int[][] ways) {

if (y > chess[0].length || x > chess.length) {
return;
}

if (visited[x][y] == false) {
ways[x][y] = level;
visited[x][y] = true;
}

if (level == (chess.length*chess[0].length-1)) {
b = false;
show(ways);
}

for (int i = 0; i < chess.length; i++) {
for (int j = 0; j < chess[i].length; j++) {

if (j == y + 1 && i == x + 2 && j <= chess[0].length
&& i <= chess.length) {
if (visited[i][j] == false) {

dfs(level + 1, i, j, chess, visited, ways);
visited[i][j] = false;
}
}

if (j == y + 2 && i == x + 1 && j <= chess[0].length
&& i <= chess.length) {
if (visited[i][j] == false) {
dfs(level + 1, i, j, chess, visited, ways);
visited[i][j] = false;
}
}

if (j == y - 1 && i == x + 2 && j <= chess[0].length
&& i <= chess.length) {
if (visited[i][j] == false) {
dfs(level + 1, i, j, chess, visited, ways);
visited[i][j] = false;
}
}

if (j == y - 2 && i == x + 1 && j <= chess[0].length
&& i <= chess.length) {
if (visited[i][j] == false) {
dfs(level + 1, i, j, chess, visited, ways);
visited[i][j] = false;
}
}

if (j == y + 1 && i == x - 2 && j <= chess[0].length
&& i <= chess.length) {
if (visited[i][j] == false) {
dfs(level + 1, i, j, chess, visited, ways);
visited[i][j] = false;
}
}

if (j == y + 2 && i == x - 1 && j <= chess[0].length
&& i <= chess.length) {
if (visited[i][j] == false) {
dfs(level + 1, i, j, chess, visited, ways);
visited[i][j] = false;
}
}

if (j == y - 1 && i == x - 2 && j <= chess[0].length
&& i <= chess.length) {
if (visited[i][j] == false) {
dfs(level + 1, i, j, chess, visited, ways);
visited[i][j] = false;
}
}

if (j == y - 2 && i == x - 1 && j <= chess[0].length
&& i <= chess.length) {
if (visited[i][j] == false) {
dfs(level + 1, i, j, chess, visited, ways);
visited[i][j] = false;
}
}

}
}
}

public static void show(int[][] ways) {
for (int i = 0; i < ways.length; i++) {
for (int j = 0; j < ways[i].length; j++) {
System.out.print(ways[i][j] + "  ");
}
System.out.println();
}
}
}

题目大意]:

给定一个p*q国际象棋棋盘，问马（骑士Knight）能否从某个点开始以跳马规则(横一步竖两步或横两步竖一步)将整个棋盘遍历；要求每个格子只能跳过一次，能的话，打印路线；

Sample Input:（先输入行，在输入列）

1 1

2 3

4 3

Sample Output:

0

不存在这样的点

0 7 2 （4 3的两种路线）

3 10 5

6 1 8

9 4 11

0 11 2

3 8 5

6 1 10

9 4 7