POJ 2296 Map Labeler

题目链接

题意：

坐标轴上有N个点。要在每一个点上贴一个正方形，这个正方形的横竖边分别和x，y轴平行，而且要使得点要么在正方形的上面那条边的中点，或者在以下那条边的中点。而且随意两个点的正方形都不重叠（能够重边）。问正方形最大边长能够多少？

思路：显然的2-sat问题，注意推断两个矩形相交的地方，细节

代码：

#include <cstdio>

#include <cstring>

#include <cstdlib>

#include <vector>

#include <algorithm>

using namespace std;

const int MAXNODE = 205;

struct TwoSet {

	int n;

	vector<int> g[MAXNODE * 2];

	bool mark[MAXNODE * 2];

	int S[MAXNODE * 2], sn;

	void init(int tot) {

		n = tot * 2;

		for (int i = 0; i < n; i += 2) {

			g[i].clear();

			g[i^1].clear();

		}

		memset(mark, false, sizeof(mark));

	}

	void add_Edge(int u, int uval, int v, int vval) {

		u = u * 2 + uval;

		v = v * 2 + vval;

		g[u^1].push_back(v);

		g[v^1].push_back(u);

	}

	void delete_Edge(int u, int uval, int v, int vval) {

		u = u * 2 + uval;

		v = v * 2 + vval;

		g[u^1].pop_back();

		g[v^1].pop_back();

	}

	bool dfs(int u) {

		if (mark[u^1]) return false;

		if (mark[u]) return true;

		mark[u] = true;

		S[sn++] = u;

		for (int i = 0; i < g[u].size(); i++) {

			int v = g[u][i];

			if (!dfs(v)) return false;

		}

		return true;

	}

	bool solve() {

		for (int i = 0; i < n; i += 2) {

			if (!mark[i] && !mark[i + 1]) {

				sn = 0;

				if (!dfs(i)){

					for (int j = 0; j < sn; j++)

						mark[S[j]] = false;

					sn = 0;

					if (!dfs(i + 1)) return false;

				}

			}

		}

		return true;

	}

} gao;

const int N = 105;

int t, n;

struct Point {

	int x, y;

	void read() {

		scanf("%d%d", &x, &y);

		x *= 2;

	}

} p[N];

bool judge(int len) {

	gao.init(n);

	for (int i = 0; i < n; i++) {

		for (int j = i + 1; j < n; j++) {

			if (p[i].x + len <= p[j].x - len || p[j].x + len <= p[i].x - len) continue;

			for (int x = 0; x < 2; x++) {

				for (int y = 0; y < 2; y++) {

					int y1, y2, y3, y4;

					if (x == 0) {

						y1 = p[i].y - len;

						y2 = p[i].y;

					} else {

						y1 = p[i].y;

						y2 = p[i].y + len;

					}

					if (y == 0) {

						y3 = p[j].y - len;

						y4 = p[j].y;

					} else {

						y3 = p[j].y;

						y4 = p[j].y + len;

					}

					if ((y1 >= y3 && y1 < y4)

							|| (y2 > y3 && y2 <= y4)

							|| (y3 >= y1 && y3 < y2)

							|| (y3 > y2 && y4 <= y2))

						gao.add_Edge(i, x, j, y);

				}

			}

		}

	}

	return gao.solve();

}

int main() {

	scanf("%d", &t);

	while (t--) {

		scanf("%d", &n);

		for (int i = 0; i < n; i++)

			p[i].read();

			int l = 0, r = 20000;

			while (l < r) {

				int mid = (l + r) / 2;

				if (judge(mid)) l = mid + 1;

				else r = mid;

			}

			printf("%d\n", l - 1);

	}

	return 0;

}