1006.cpp

我的URAL的1000 ~ 1050 的全部代码 包含WA 最后AC的程序 有2~3个比较难的是MAIGO的程序

//找到一个框框之后，把它拿掉，被它覆盖的地方标上?。?可以代表任何东西。当我们只剩下?和.的时候，工作完成！
// hint : all frame are square  x == y;
#include <iostream>
using namespace std;
const int maxX = 50;
const int maxY = 20;
const char c1 = 218 , c2 = 191, c3 = 192, c4 = 217, cv = 179, ch = 196, any = '?';	//提交的时候注意要改
char map[maxY][maxX];

int ans[2000][3];	// total sybol square length
int cnt,totS;

int min(int a, int b){ return a<b?a:b; }

void readIn()
{
	int i, j;
	totS = 0;
	for( j = 0; j < maxY; j++ )
	{
		for( i = 0; i < maxX; i++ )
		{
			cin >> map[j][i];
			if( map[j][i] != '.' ) totS++;
		}
	}
}

bool ok( int x, int y, int a )
{
	if( map[y][x] != c1 && map[y][x] != any )     	  return false;
	if( map[y][x+a] != c2 && map[y][x+a] != any )	  return false;
	if( map[y+a][x] != c3 && map[y+a][x] != any ) 	  return false;
	if( map[y+a][x+a] != c4 && map[y+a][x+a] != any ) return false;
	return true;
}

void clear( int x, int y )
{
	if( map[y][x] != '.' && map[y][x] != any ) 
		{totS--;	map[y][x] = any;}
}
bool check( int x, int y , int a )
{
	int i;
	for( i = x+1; i< x+a; i++ )
	{
		if( map[y][i] != ch && map[y][i] != any )return false;
		if( map[y+a][i] != ch && map[y+a][i] != any )return false;
	}
	for( i = y+1; i < y+a; i++ )
	{
		if( map[i][x] != cv && map[i][x] != any )return false;
		if( map[i][x+a] != cv && map[i][x+a] != any )return false;
	}
	int t = totS;
	for( i = x; i <= x+a; i++ )
		{clear(i,y); clear(i,y+a);}
	for( i = y; i <= y+a; i++ )
		{clear(x,i); clear(x+a,i);}
	if( totS == t )return false;
	else return true;
}


void output()
{
	int i;
	cout << cnt << endl;
	for( i = cnt; i >= 1; i-- )
		cout << ans[i][0] << " " << ans[i][1] << " " << ans[i][2] <<endl;
}
int main()
{
	int i, j, l;
//	freopen("1006.in","r",stdin);
	readIn();
	while( totS > 0 )
	{
//		cout << totS <<endl;
		for( l = 1; l < maxY; l++ )
			for( i = 0; i < maxY - l; i++ )
				for( j = 0; j < maxX - l; j++ )
					if( ok(j, i, l) )
						if( check(j, i, l) )
						{
							cnt++;
						ans[cnt][0] = j, ans[cnt][1] = i, ans[cnt][2]= l+1;
						}
	}
	output();
	return 0;
}