HDU 1815， POJ 2749 Building roads

题目链接HDU

题目链接POJ

题意:

有n个牛棚， 还有两个中转站S1和S2， S1和S2用一条路连接起来。 为了使得任意牛棚两个都可以有道路联通，现在要让每个牛棚都连接一条路到S1或者S2。

有a对牛棚互相有仇恨，所以不能让他们的路连接到同一个中转站。还有b对牛棚互相喜欢，所以他们的路必须连到同一个中专站。

道路的长度是两点的曼哈顿距离。

问最小的任意两牛棚间的距离中的最大值是多少？

思路：二分距离，考虑每两个牛棚之间4种连边方式，然后根据二分的长度建立表达式，然后跑2-sat判断即可

代码：

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <vector>
#include <algorithm>
#include <cmath>
using namespace std;

const int MAXNODE = 505;

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 = 505;
int n, a, b;

struct Point {
	int x, y;
	void read() {
		scanf("%d%d", &x, &y);
	}
} s1, s2, p[N], A[N * 2], B[N * 2];

int dis(Point a, Point b) {
	int dx = a.x - b.x;
	int dy = a.y - b.y;
	return abs(dx) + abs(dy);
}

int g[N][N][4];

bool judge(int d) {
	gao.init(n);
	for (int i = 0; i < a; i++) {
		gao.add_Edge(A[i].x - 1, 0, A[i].y - 1, 0);
		gao.add_Edge(A[i].x - 1, 1, A[i].y - 1, 1);
	}
	for (int i = 0; i < b; i++) {
		gao.add_Edge(B[i].x - 1, 0, B[i].x - 1, 1);
		gao.add_Edge(B[i].x - 1, 0, B[i].y - 1, 1);
		gao.add_Edge(B[i].y - 1, 0, B[i].x -1 , 1);
		gao.add_Edge(B[i].y - 1, 0, B[i].y - 1, 1);
	}
	for (int i = 0; i < n; i++) {
		for (int j = 0; j < i; j++) {
			if (g[i][j][3] > d)
				gao.add_Edge(i, 0, j, 1);
			if (g[i][j][2] > d)
				gao.add_Edge(i, 1, j, 0);
			if (g[i][j][1] > d)
				gao.add_Edge(i, 0, j, 0);
			if (g[i][j][0] > d)
				gao.add_Edge(i, 1, j, 1);
		}
	}
	return gao.solve();
}

int main() {
	while (~scanf("%d%d%d", &n, &a, &b)) {
		s1.read(); s2.read();
		for (int i = 0; i < n; i++) {
			p[i].read();
			for (int j = 0; j < i; j++) {
				g[i][j][0] = dis(p[i], s1) + dis(p[j], s1);
				g[i][j][1] = dis(p[i], s2) + dis(p[j], s2);
				g[i][j][2] = dis(p[i], s1) + dis(p[j], s2) + dis(s1, s2);
				g[i][j][3] = dis(p[i], s2) + dis(p[j], s1) + dis(s1, s2);
			}
		}
		for (int i = 0; i < a; i++) A[i].read();
		for (int i = 0; i < b; i++) B[i].read();
		int l = 0, r = 7777777;
		if (!judge(r)) printf("-1\n");
		else {
			while (l < r) {
				int mid = (l + r) / 2;
				if (judge(mid)) r = mid;
				else l = mid + 1;
			}
			printf("%d\n", l);
		}
	}
	return 0;
}