G - 深搜 基础
Crawling in process... Crawling failed Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
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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 解析见代码代码：
/*
hdu2488 深搜，判断能否走完全图，并要求输出路径
首先是能否走完全图的判断,深搜函数加一个参数step，
来判断是否走完全图，同时用flag进行标记，方便输出两种情况
再就是路径如何保存，只需要保存每一步的x,y坐标即可，使用
一个二位组就可以保存。同时因为vis数组是以步数为标准来进行保存的
其值会随着递归回溯不断更新，始终保证是最新解,step=p*q标志着递归
成功，按照步数输出即可，注注意格式要求的是先输纵坐标后输横坐标
*/
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <stack>
#include <queue>
using namespace std;
const int maxn=100;
int vis[maxn*maxn][2];//vis二维数组，前一个参数代表是第几步，后一个参数0代表横坐标，后一个参数代表纵坐标。
int p,q,step,flag;
char maps[maxn][maxn];
int f[8][2]={{-1,-2},{1,-2},{-2,-1},{2,-1},{-2,1},{2,1},{-1,2},{1,2}};//输出要求按字典序输出，同时注意大写字母是列编号，所以方向数组应该是按照先y后x字典序开
int ans=0;
int dis[maxn][maxn];
void dfs(int x,int y,int step)
{
if(step==p*q&&flag==0)
{
cout<<"Scenario #"<<++ans<<":"<<endl;
for(int i=0;i<p*q;i++)
printf("%c%d",‘A‘+vis[i][1],vis[i][0]+1);
flag=1;
cout<<endl<<endl;//输出格式要求
return;
}
for(int i=0;i<8;i++)
{
int a=x+f[i][0];
int b=y+f[i][1];
if(a>=0&&a<p&&b>=0&&b<q&&!dis[a][b])
{
dis[a][b]=1;
vis[step][0]=a;
vis[step][1]=b;
dfs(a,b,step+1);
dis[a][b]=0;//回溯时该点状态恢复
if(flag) return;//相当于一个剪枝操作，找到就返回，大大提高了程序工作效率
}
}
}
int main()
{
int n;
cin>>n;
while(n--)
{
memset(dis,0,sizeof(dis));
cin>>p>>q;
flag=0;
dis[0][0]=1;//标记数组，避免重复搜索
vis[0][0]=0,vis[0][1]=0;
dfs(0,0,1);
if(!flag)//用flag来判断最终是否走完全图
{
cout<<"Scenario #"<<++ans<<":"<<endl;
cout<<"impossible"<<endl<<endl;
}
}
return 0;
}