HDU 1815， POJ 2749 Building roads

pid=1815" target="_blank" style="">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。

为了使得随意牛棚两个都能够有道路联通，如今要让每一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。

问最小的随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，然后依据二分的长度建立表达式，然后跑2-sat推断就可以

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;

}