poj 2488 A Knight's Journey(简单dfs)

而且起点必须是A1。

#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <set>
#include <map>
#include <vector>
#include <math.h>
#include <string.h>
#include <queue>
#include <string>
#define LL long long
#define _LL __int64
#define eps 1e-8

using namespace std;
const int INF = 0x3f3f3f3f;
const int maxn = 26;
const int maxm = 1010;
int dir[8][2] = {{-1,-2},{1,-2},{-2,-1},{2,-1},{-2,1},{2,1},{-1,2},{1,2}};
int n,m;
int vis[maxn][maxn];
struct node
{
	int x;
	int y;
}ans[30];

bool dfs(int x, int y ,int step)
{
	vis[x][y] = 1;
	ans[step] = ((struct node){x,y});

	if(step == n*m)
		return true; //递归边界

	for(int d = 0; d < 8; d++)
	{
		int xx = x + dir[d][0];
		int yy = y + dir[d][1];
		if(xx >= 1 && xx <= n && yy >= 1 && yy <= m && !vis[xx][yy])
			if(dfs(xx,yy,step+1))
				return true;
	}
	vis[x][y] = 0; //走到这一步，说明（x,y）走不通，所以要重新标记为0
	return false;
}

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

	for(int item = 1; item <= test; item++)
	{
		scanf("%d %d",&n,&m);

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

		printf("Scenario #%d:\n",item);

		if(dfs(1,1,1))
		{
			for(int i = 1; i <= n*m; i++)
				printf("%c%d",ans[i].y-1+'A',ans[i].x);
			printf("\n\n");
		}
		else printf("impossible\n\n");

	}
	return 0;
}