? ? ? 在國際象棋的棋盤（8行×8列）上放置一個馬，按照“馬走日字”的規(guī)則，馬要遍歷棋盤，即到達棋盤上的每一格，并且每格只到達一次。例如，下圖給出了騎士從坐標（1,5）出發(fā)，游歷棋盤的一種可能情況。

【例1】騎士游歷問題。

? ? ? ?編寫一個程序，對于給定的起始位置（x0,y0），探索出一條路徑，沿著這條路徑騎士能遍歷棋盤上的所有單元格。

? ? ? ?（1）編程思路。

? ? ? ? 采用深度優(yōu)先搜索進行路徑的探索。深度優(yōu)先搜索用遞歸描述的一般框架為：

? ? void? dfs(int deep)????? //? 對deep層進行搜索

??? {

????????? if? (符合某種要求||已經(jīng)不能再搜了)

???????? {

?????????????? 按要求進行一些處理，一般為輸出;

?????????????? return ;

???????? }

???????? if?? (符合某種條件且有地方可以繼續(xù)搜索的)?? // 這里可能會有多種情況，可能要循環(huán)什么的

??????? {

? ? ? ? ? ? ?vis[x][y]=1;?? ????????????????????//? 表示結(jié)點(x，y)已訪問到

? ? ? ? ? ? ?dfs(deep+1);??????????????????? //? 搜索下一層

? ? ? ? ? ? ?vis[x][y]=0;????????????????????? // 改回來，表示結(jié)點(x，y)以后可能被訪問
??????? }
??? }?

? ? ? 定義數(shù)組int vis[10][10]記錄騎士走到的步數(shù)，vis[x][y]=num表示騎士從起點開始走到坐標為（x,y）的格子用了num步（設(shè)起點的步數(shù)為1）。初始時vis數(shù)組元素的值全部為0。

?（2）源程序。

#include <stdio.h>

#include <stdlib.h>

int N,M;

int vis[10][10]={0};

// 定義馬走的8個方向

int dir_x[8] = {-1,-2,-2,-1,1,2,2,1};

int dir_y[8] = {2,1,-1,-2,-2,-1,1,2};

void print()

{

? ? ? ?int i,j;

? ? ? ?for(i=0; i<N; i++)

? ? ? ?{

? ? ? ? ? ?for(j=0; j<M; j++)

? ? ? ? ? ? ? ?printf("%3d ",vis[i][j]);

? ? ? ? ? ?printf("\n");

? ? ? ? }

}

void DFS(int cur_x,int cur_y,int step)

{

?? if(step==N*M+1 )

?? {

? ? ? ? print();

??????? exit(1);

?? }?

?? int next_x,next_y;

?? for(int i=0; i<8; i++)

?? {

???? next_x = cur_x+dir_x[i];

???? next_y = cur_y+dir_y[i];

???? if (next_x<0 || next_x>=N || next_y<0 || next_y>=M || vis[next_x][next_y]!=0)

? ? ? ? ? continue;

???? vis[next_x][next_y] = step;

???? DFS(next_x,next_y,step+1);

???? vis[next_x][next_y] = 0;

?? }

}

int main()

{

?? printf("請輸入棋盤的行數(shù)和列數(shù)(均小于10)：");

?? scanf("%d %d",&N,&M);

?? printf("請輸入出發(fā)點坐標：(0—%d,0-%d)：",N-1,M-1);

?? int x0,y0;

?? scanf("%d%d",&x0,&y0);

?? vis[x0][y0] = 1;

?? DFS(x0,y0,2);

?? return 0;

}?

? ? ? ?（3）運行效果。

? ???

?【例2】A Knight's Journey（POJ 2488）

Description
Background
The knight is getting bored of seeing the same black and white squares again and again and has decided to make a journey around the world. Whenever a knight moves, it is two squares in one direction and one square perpendicular to this. The world of a knight is the chessboard he is living on. Our knight lives on a chessboard that has a smaller area than a regular 8 * 8 board, but it is still rectangular. Can you help this adventurous knight to make travel plans?
Problem
Find a path such that the knight visits every square once. The knight can start and end on any square of the board.
Input
The input begins with a positive integer n in the first line. The following lines contain n test cases. Each test case consists of a single line with two positive integers p and q, such that 1 <= p * q <= 26. This represents a p * q chessboard, where p describes how many different square numbers 1, . . . , p exist, q describes how many different square letters exist. These are the first q letters of the Latin alphabet: A, . . .
Output
The output for every scenario begins with a line containing "Scenario #i:", where i is the number of the scenario starting at 1. Then print a single line containing the lexicographically first path that visits all squares of the chessboard with knight moves followed by an empty line. The path should be given on a single line by concatenating the names of the visited squares. Each square name consists of a capital letter followed by a number.
If no such path exist, you should output impossible on a single line.
Sample Input
3
1 1
2 3
4 3
Sample Output
Scenario #1:
A1

Scenario #2:
impossible

Scenario #3:
A1B3C1A2B4C2A3B1C3A4B2C4

? ? ? （1）編程思路。

? ? ? ?同樣用深度優(yōu)先搜索。但由于題目要輸出字典序最小的，所以遍歷時8個方向的偏移組合順序為：{-2,-1}, {-2,1}, {-1,-2}, {-1,2}, {1,-2}, {1,2}, {2,-1}, {2,1}。

? ? ? （2）源程序。

#include<stdio.h>??

int dir_x[8] = {-2,-2,-1,-1, 1, 1, 2, 2};

int dir_y[8] = {-1, 1,-2, 2,-2, 2,-1, 1};

int vis[27][27];?

int len,x,y;?

bool flag;?

struct Node?

{?

??? int x,y;?

}node[1000];?

void DFS(int cur_x,int cur_y)?

{?

??? if(len==x*y)?

??? {?

??????? flag=true;?

??????? return ;?

??? }?

??? for(int i=0; i<8; i++)?

??? {?

??????? int next_x=cur_x+dir_x[i];?

??????? int next_y=cur_y+dir_y[i];?

??????? if(next_x>0 && next_x<=x && next_y>0 && next_y<=y && vis[next_x][next_y]!=1)?

? ??????{?

??????????? node[len].x=next_x;?

??????????? node[len].y=next_y;?

??????????? vis[next_x][next_y]=1;?

??????????? ++len;

??????????? DFS(next_x,next_y);?

??????????? if(len==x*y)?

??????????? {?

??????????????? flag=true;?

??????????????? return ;?

??????????? }?

??????????? --len;?

??????????? vis[next_x][next_y]=0;?

??????? }?

??? }?

}?

int main()?

{?

??? int nCase;?

??? int n,i,j;?

??? scanf("%d",&nCase);?

??? for(n=1; n<=nCase; n++)?

??? {?

??????? flag=false;?

??????? len=0;

? ? ? ? for (i=0;i<27;i++)

? ? ? ? ? ?for (j=0;j<27;j++)

?????????????? vis[i][j]=0;?

??????? node[0].x=1;?

??????? node[0].y=1;?

??????? vis[1][1]=1;?

??????? scanf("%d%d",&y,&x);?

??????? ++len;?

??????? DFS(1,1);?

??????? printf("Scenario #%d:\n",n);?

??????? if(flag==false)?

??????? {?

??????????? printf("impossible\n\n");

??????????? continue;?

??????? }?

??????? for(i=0; i<len; i++)?

??????? {?

??????????? printf("%c%d",(node[i].x-1)+'A',node[i].y);?

??????? }?

??????? printf("\n\n");

? ? ? ??? ? }?

??? return 0;?

}

?

轉(zhuǎn)載于:https://www.cnblogs.com/cs-whut/p/11153529.html

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