A - A Knight's Journey （poj2488）(dfs经典)
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2022-03-31 10:05:09
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A Knight's Journey
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 27258 | Accepted: 9295 |
Description
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.
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
大致题意：给出一个p行q列的国际棋盘，马可以从任意一个格子开始走，问马能否不重复的走完所有的棋盘。如果可以，输出按字典序排列最小的路径。打印路径时，列用大写字母表示（A表示第一列），行用阿拉伯数字表示（从1开始），先输出列，再输出行。
分析：如果马可以不重复的走完所有的棋盘，那么它一定可以走到A1这个格子。所以我们只需从A1这个格子开始搜索，就能保证字典序是小的；除了这个条件，我们还要控制好马每次移动的方向，控制方向时保证字典序最小（即按照下图中格子的序号搜索）。控制好这两个条件，直接从A1开始深搜就行了。
代码：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<deque>
#include<map>
using namespace std;
int p,q;
int sum;
bool g[30][30];
//int dx[8]={-1,-1,-2,-2,1,1,2,2};
//int dy[8]={-2,2,-1,1,-2,2,-1,1};
int dx[8]={-1,1,-2,2,-2,2,-1,1};
int dy[8]={-2,-2,-1,-1,1,1,2,2};
bool find1;
struct point
{
int x,y;
point(int xx,int yy):x(xx),y(yy){}
};
deque<point> dq;
void dfs(int x,int y,int step)
{
if(step==sum)
{
find1=true;
}
if(find1)
return;
if(x<1||x>p||y<1||y>q)
return;
int i;
for(i=0;i<8;i++)
{
int xx=dx[i]+x;
int yy=dy[i]+y;
if(x<1||x>p||y<1||y>q)
continue;
if(g[xx][yy]==0&&!find1)
{
g[xx][yy]=1;
dq.push_back(point(xx,yy));
dfs(xx,yy,step+1);
if(find1)
return;
dq.pop_back();
g[xx][yy]=0;
}
}
}
int main(void)
{
//map<pair<int,int>,string> mp;
int t;
scanf("%d",&t);
for(int i=1;i<=t;i++)
{
scanf("%d %d",&p,&q);
//if(mp.find(make_pair(p,q))!=mp.end())
//{
// printf("Scenario #%d:\n",i);
// cout<<mp[make_pair(p,q)]<<endl<<endl;
// continue;
//}
sum=p*q;
memset(g,0,sizeof(g));
find1=0;
// int k,j;
// for(j=1;j<=p;j++)
// {
// for(k=1;k<=q;k++)
// {
// dq.clear();
// g[k][j]=1;
// dq.push_back(point(k,j));
// dfs(k,j,0);
// g[k][j]=0;
// if(find1)
// break;
// }
// if(find1)
// break;
// }
dq.clear();
g[1][1]=1;
dq.push_back(point(1,1));
dfs(1,1,1);
printf("Scenario #%d:\n",i);
if(find1)
{
//string a="";
while(dq.size())
{
point now=dq.front();
dq.pop_front();
//a+=('A'+(now.y-1));
//a+=(now.x+'0');
printf("%c%d",'A'+(now.y-1),now.x);
}
//mp.insert(make_pair(make_pair(p,q),a));
printf("\n");
}
else
{
printf("impossible\n");
}
if(i!=t)
printf("\n");
}
return 0;
}代码：（优化了用时却多了）
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<deque>
#include<map>
using namespace std;
int p,q;
int sum;
bool g[30][30];
//int dx[8]={-1,-1,-2,-2,1,1,2,2};// !!! lexicographically !!!
//int dy[8]={-2,2,-1,1,-2,2,-1,1};
int dx[8]={-1,1,-2,2,-2,2,-1,1};
int dy[8]={-2,-2,-1,-1,1,1,2,2};
bool find1;
struct point
{
int x,y;
point(int xx,int yy):x(xx),y(yy){}
};
deque<point> dq;
void dfs(int x,int y,int step)
{
if(step==sum)
{
find1=true;
dq.pop_back();
}
if(find1)
return;
if(x<1||x>p||y<1||y>q)
return;
int i;
for(i=0;i<8;i++)
{
int xx=dx[i]+x;
int yy=dy[i]+y;
if(x<1||x>p||y<1||y>q)
continue;
if(g[xx][yy]==0)
{
g[xx][yy]=1;
dq.push_back(point(xx,yy));
dfs(xx,yy,step+1);
if(find1)
return;
dq.pop_back();
g[xx][yy]=0;
}
}
}
int main(void)
{
map<pair<int,int>,string> mp;
int t;
scanf("%d",&t);
for(int i=1;i<=t;i++)
{
scanf("%d %d",&p,&q);
if(mp.find(make_pair(p,q))!=mp.end())
{
printf("Scenario #%d:\n",i);
cout<<mp[make_pair(p,q)]<<endl<<endl;
continue;
}
sum=p*q;
memset(g,0,sizeof(g));
find1=0;
dq.clear();
g[1][1]=1;
dfs(1,1,0);
dq.push_front(point(1,1));
printf("Scenario #%d:\n",i);
if(find1)
{
string a="";
while(dq.size())
{
point now=dq.front();
dq.pop_front();
a+=('A'+(now.y-1));
a+=(now.x+'0');
printf("%c%d",'A'+(now.y-1),now.x);
}
mp.insert(make_pair(make_pair(p,q),a));
printf("\n");
}
else
{
printf("impossible\n");
}
if(i!=t)
printf("\n");
}
return 0;
}代码：（别人的）
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int path[88][88], vis[88][88], p, q, cnt;
bool flag;
int dx[8] = {-1, 1, -2, 2, -2, 2, -1, 1};
int dy[8] = {-2, -2, -1, -1, 1, 1, 2, 2};
bool judge(int x, int y)
{
if(x >= 1 && x <= p && y >= 1 && y <= q && !vis[x][y] && !flag)
return true;
return false;
}
void DFS(int r, int c, int step)
{
path[step][0] = r;
path[step][1] = c;
if(step == p * q)
{
flag = true;
return ;
}
for(int i = 0; i < 8; i++)
{
int nx = r + dx[i];
int ny = c + dy[i];
if(judge(nx,ny))
{
vis[nx][ny] = 1;
DFS(nx,ny,step+1);
vis[nx][ny] = 0;
}
}
}
int main()
{
int i, j, n, cas = 0;
scanf("%d",&n);
while(n--)
{
flag = 0;
scanf("%d%d",&p,&q);
memset(vis,0,sizeof(vis));
vis[1][1] = 1;
DFS(1,1,1);
printf("Scenario #%d:\n",++cas);
if(flag)
{
for(i = 1; i <= p * q; i++)
printf("%c%d",path[i][1] - 1 + 'A',path[i][0]);
}
else
printf("impossible");
printf("\n");
if(n != 0)
printf("\n");
}
return 0;
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