hdu2488 dfs

Posted 2020-07-22 当蜗牛有了理想

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;
}