JZOJ 6799. 【2014广州市选day2】game

题目

思路

呵呵，正解并不是什么神奇的方法

而是最原始的最粗暴的最有用的最万能的————搜索

依题模拟即可

\(Code\)

#include<cstdio>
#include<cstring>
using namespace std;

const int N = 35;
int n , m , mp[N][N] , vis[N][N] , ans , up;
char s[N];

void dfs(int x , int y , int s , int num)
{
	if (s >= ans) return;
	if (num == up){ans = s; return;}
	vis[x][y] = 1;
	int l , r;

	l = x , r = y;
	while (r + 1 <= m && !mp[l][r + 1] && !vis[l][r + 1]) vis[l][++r] = 1;
	if (r != y)
	{
		dfs(l , r , s + 1 , num + r - y);
		for(register int i = y + 1; i <= r; i++) vis[x][i] = 0;
	}

	l = x , r = y;
	while (r - 1 > 0 && !mp[l][r - 1] && !vis[l][r - 1]) vis[l][--r] = 1;
	if (r != y)
	{
		dfs(l , r , s + 1 , num + y - r);
		for(register int i = r; i < y; i++) vis[x][i] = 0;
	}

	l = x , r = y;
	while (l + 1 <= n && !mp[l + 1][r] && !vis[l + 1][r]) vis[++l][r] = 1;
	if (l != x)
	{
		dfs(l , r , s + 1 , num + l - x);
		for(register int i = x + 1; i <= l; i++) vis[i][y] = 0;
	}

	l = x , r = y;
	while (l - 1 > 0 && !mp[l - 1][r] && !vis[l - 1][r]) vis[--l][r] = 1;
	if (l != x)
	{
		dfs(l , r , s + 1 , num + x - l);
		for(register int i = l; i < x; i++) vis[i][y] = 0;
	}
	vis[x][y] = 0;
}

int main()
{
	freopen("game.in" , "r" , stdin);
	freopen("game.out" , "w" , stdout);
	scanf("%d%d" , &n , &m);
	up = n * m;
	for(register int i = 1; i <= n; i++)
	{
		scanf("%s" , s);
		for(register int j = 0; j < m; j++)
		if (s[j] == '*')  mp[i][j + 1] = 1 , --up;
	}
	ans = 2147483647;
	for(register int i = 1; i <= n; i++)
		for(register int j = 1; j <= m; j++)
		if (!mp[i][j])
		{
			memset(vis , 0 , sizeof vis);
			dfs(i , j , 0 , 1);
		}
	printf("%d" , ans == 2147483647 ? -1 : ans);
}