bzoj1052: [HAOI2007]覆盖有关问题

先二分边长，然后按这样的策略判断：正方形肯定是选边上的四个角，枚举第一块在哪个角，再算一边四个角，再枚举第二块，判断剩下的能不能被第三块覆盖

const int N = 20010;
pair<int, int> Data[N];
int n;

inline void Input() {
	scanf("%d", &n);
	For(i, 1, n) scanf("%d%d", &Data[i].ft, &Data[i].sd);
}

int Mark[N];

inline bool Solve(int R, int Dep) {
	int Up = -MIT, Down = MIT, Left = MIT, Right = -MIT, flag = 0;
	For(i, 1, n)
		if(!Mark[i]) {
			flag = 1;
			int X = Data[i].ft, Y = Data[i].sd;
			Up = max(Up, Y), Down = min(Down, Y),
			Left = min(Left, X), Right = max(Right, X);
		}
	if(!flag) return 1;
	
	if(Dep > 2) return R >= max(Up - Down, Right - Left);
	else {
		int Ox1, Ox2, Oy1, Oy2;
		For(type, 1, 4) {
			if(type == 1) Ox1 = Left, Ox2 = Left + R, Oy1 = Down, Oy2 = Down + R;
			else if(type == 2) Ox1 = Left, Ox2 = Left + R, Oy1 = Up - R, Oy2 = Up;
			else if(type == 3) Ox1 = Right - R, Ox2 = Right, Oy1 = Down, Oy2 = Down + R;
			else Ox1 = Right - R, Ox2 = Right, Oy1 = Up - R, Oy2= Up;
			
			// mark
			For(i, 1, n)
				if(!Mark[i]) {
					pair<int, int> *Dat = &Data[i];
					if(Dat->ft >= Ox1 && Dat->ft <= Ox2 &&
						Dat->sd >= Oy1 && Dat->sd <= Oy2) Mark[i] = Dep;
				}
			if(Solve(R, Dep + 1)) return 1;
			For(i, 1, n)
				if(Mark[i] == Dep) Mark[i] = 0;
		}
	}
	
	return 0;
}
inline void Solve() {
	// get MaxDis
	int Up = -MIT, Down = MIT, Left = MIT, Right = -MIT;
	For(i, 1, n) {
		int X = Data[i].ft, Y = Data[i].sd;
		Up = max(Up, Y), Down = min(Down, Y),
		Left = min(Left, X), Right = max(Right, X);
	}
	
	int mid, ans = INF, lef = 0, rig = max(Up - Down, Left - Right);
	while(lef <= rig) {
		mid = (lef + rig) >> 1;
		clr(Mark, 0);
		if(Solve(mid, 1)) rig = (ans = mid) - 1;
		else lef = mid + 1;
	}
	
	printf("%d\n", ans);
}

int main() {
    #ifndef ONLINE_JUDGE
    SETIO("1052");
    #endif
    Input();
    Solve();
    return 0;
}