POJ 2488 A Knight's Journey (DFS)

輸入一個棋盤 ( p × q )，從任意點開始，每次走日子格，求能走遍每一個點的序列，以Sample中給出的樣式按字典序輸出，若是得不到這樣的序列輸出impossible

因爲要按字典序輸出，因此優先訪問座標較小的點，第一次獲得的序列就是所求結果

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>

using namespace std;

int const maxn = 30;

bool vis[maxn][maxn];
int p, q;
bool flag = false;

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

typedef struct node
{
	int x;
	int y;
}coordinate;

bool judge()
{
	for( int i = 0; i < p; i++ )
		for( int j = 0; j < q; j++ )
			if( !vis[i][j] )
				return false;
	return true;
}

vector<coordinate> path;

void dfs( int x, int y )
{
    if( flag )
        return;
	if( judge() && !flag ) {
        printf( "A1" );
		for( int i = 0; i < path.size(); i++ ){
			printf( "%c", path[i].y + 65 );
			printf( "%d", path[i].x + 1 );
		}
		puts("");
		flag = true;
		return;
	}
	for( int i = 0; i < 8; i++ ) {
		int next_x = x + steps[i][0];
		int next_y = y + steps[i][1];
		if( next_x >= 0 && next_x < p && next_y >= 0 && next_y < q && !vis[next_x][next_y] ) {

			coordinate tmp;
			tmp.x = next_x;
			tmp.y = next_y;

			path.push_back( tmp );
			vis[next_x][next_y] = 1;

			dfs( next_x, next_y );

			vis[next_x][next_y] = 0;
			path.pop_back();
		}
	}

}

int main()
{
	int n; scanf( "%d", &n );

	for( int i = 0; i < n; i++ ) {
		scanf( "%d%d", &p, &q );

		flag = false;

		printf( "Scenario #%d:\n", i + 1 );

		memset( vis, 0, sizeof( vis ) );

		path.clear();
		vis[0][0] = 1;
		dfs( 0, 0 );
		if( !flag )
            printf( "impossible\n" );
        puts("");

	}
	return 0;
}