TopCoder——SRM 521 DIV 2

老外的眼神真好，咱英语水平还有待提高！

Score 250. 枚举分割点，取最小值. ( 比赛时挂掉了，好郁闷，思考问题还是欠缺严谨呐 )

class RedAndGreen
{
public:
	int minPaints(string row) {
		int Len = (int)row.length();
		int ans = 0x7fffffff;
		for(int i = 0; i <= Len; ++i) {
			int tot = 0;
			for(int j = 0; j < i; ++j)
				if( row[j] == 'G' ) ++tot;
			for(int j = i; j < Len; ++j)
				if( row[j] == 'R' ) ++tot;
			ans = min(ans, tot);
		}
		return ans;
	}
	
};

Score 500. 当右括号比左括号数目多的时候，表明需要添加二者差值个左括号，然后重新计数，最后加上需要添加的右括号数目.

class MissingParentheses 
{
public:

	int countCorrections(string par) {
		int Len = (int)par.length();
		int ans = 0, L = 0, R = 0;
		for(int i = 0; i < Len; ++i) {
			if( par[i] == '(' ) ++L;
			else ++R;
			
			if( R > L ) {
				ans += (R - L);
				L = R = 0;
			}
		}
		return ans + (L - R);
	}
	
	
};

Score 1000. 需要离散化预处理，枚举合法的矩形<(x, y), (x + n, y + n)>，不知道枚举左下脚标有什么要求 (x,y)，猜测会在(_x, _y) 的基础上，可能会是

1. 保持不变

2. 增量为 +1  /  -1

3. 增量为 +n  /  -n

4. 增量为 + ( n + 1 )   /  - ( n + 1 )

反正数据量不大，试试再说 —— Well, Passed！

class SquaredSubsets 
{
public:
	bool inSide(int &x, int &y, int &_x, int &_y, int &n) {
		return (x >= _x) && (x <= _x + n) && (y >= _y) && (y <= _y + n);
	}

	long long countSubsets(int n, vector <int> x, vector <int> y) {
		int Len = (int)x.size();
		n <<= 0x1;
		for(int i = 0; i < Len; ++i) x[i] <<= 0x1, y[i] <<= 0x1;
		
		vector<int> _x, _y;
		for(int i = 0; i < Len; ++i) {
			_x.push_back( x[i] );
			_x.push_back( x[i] + 1 );
			_x.push_back( x[i] + n );
			_x.push_back( x[i] + 1 + n );
			_x.push_back( x[i] - 1 );
			_x.push_back( x[i] - n );
			_x.push_back( x[i] - 1 - n );
			
			_y.push_back( y[i] );
			_y.push_back( y[i] + 1 );
			_y.push_back( y[i] + n );
			_y.push_back( y[i] + 1 + n );
			_y.push_back( y[i] - 1 );
			_y.push_back( y[i] - n );
			_y.push_back( y[i] - 1 - n );
		}
		
		int _Len = (int)_x.size();
		
		set<long long> d;
		
		long long v;
		for(int i = 0, j, k, tx, ty; i < _Len; ++i) {
			for(j = 0; j < _Len; ++j) {
				tx = _x[i], ty = _y[j];
				v = 0;
				for(k = 0; k < Len; ++k) 
					if( inSide(x[k], y[k], tx, ty, n) ) v |= (1LL << k);
				if( v ) d.insert(v);
			}
		}
		
		return (long long)d.size();
	}
};

但是，还是需要再花时间弄清楚解题思路，盲猜毕竟不是长久之道！