poj 2488 dfs+回溯

A Knight’s Journey

Time Limit: 1000MS Memory Limit: 65536K

Total Submissions: 36725 Accepted: 12472

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

直接搜回溯就好，水题

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <queue>
#include <stack>
//#define LOCAL
using namespace std;

typedef struct{
char sc;
int si;
}Point;

typedef struct{
int xx;
int yy;
}POint;

int G[100][100];
int m,n;
stack<Point>sta;
POint move[] = {-1,-2,1,-2,-2,-1,2,-1,-2,1,2,1,-1,2,1,2};

bool IsTrue(int x,int y){
if(x>=1&&x<=m&&y>=1&&y<=n&&!G[x][y]) return true;
return false;
}

bool IsT(void){
int i,j;
for(i = 1;i<=m;i++)
for(j = 1;j<=n;j++)
if(!G[i][j]) return false;
return true;
}

bool dfs(int x,int y){
int i;
Point term;
term.sc = 'A' + y - 1;
term.si = x;
G[x][y] = 1;
if(IsT()){
sta.push(term);
return true;
}
for(i = 0;i<8;i++){
int tx = x + move[i].xx,ty = y + move[i].yy;
if(IsTrue(tx,ty))
if(dfs(tx,ty)){
sta.push(term);
return true;
}

}
G[x][y] = 0;
return false;
}

int main()
{
#ifdef LOCAL
freopen("in.txt","r",stdin);
freopen("out1.txt","w",stdout);
#endif // LOCAL
int N;
int CNT = 1;
scanf("%d",&N);
while(N--){
scanf("%d%d",&m,&n);
memset(G,0,sizeof(G));
while(!sta.empty()) sta.pop();
bool ok = dfs(1,1);
printf("Scenario #%d:\n",CNT++);
if(ok){
while(!sta.empty()){
Point term = sta.top();
sta.pop();
printf("%c%d",term.sc,term.si);
}
printf("\n\n");
}else
printf("impossible\n\n");
}

return 0;
}