UVA 10319 - Manhattan

题目链接

题意：一个城市，有南北和东西街道。每种街道都是单行道，如今给定几个起点和终点。要求起点和终点必须最多转一次弯能够到达，问能否够满足全部的起点终点

思路：2-set,对于东西。南北街道，向西(北)为false，向东(南)为true，对于每一个起点终点，肯定是走坐标相应的那4条街道，表示出来是(s1 & a2) | (s2 & a1)，能够转化成(s1 | s2) & (s1 | a1) & (a2 | v2) & (a2 | a1)相应2-set4条边，注意假设横坐标或纵坐标同样，仅仅要建一条边(s, s)就可以。这样建图，进行2-set判定就可以

代码：

#include <cstdio>

#include <cstring>

#include <cstdlib>

#include <vector>

#include <algorithm>

using namespace std;

const int MAXNODE = 2005;

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);

	}

	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;

int n, S, A, m;

int main() {

	scanf("%d", &n);

	while (n--) {

		scanf("%d%d%d", &S, &A, &m);

		gao.init(S + A);

		int s1, a1, s2, a2;

		while (m--) {

			scanf("%d%d%d%d", &s1, &a1, &s2, &a2);

			s1--; a1--; s2--; a2--;

			if (s1 == s2 && a1 == a2) continue;

			a1 += S;

			a2 += S;

			if (s1 == s2) gao.add_Edge(s1, (a1 < a2), s2, (a1 < a2));

			else if (a1 == a2) gao.add_Edge(a1, (s1 < s2), a2, (s1 < s2));

			else {

				gao.add_Edge(s1, (a1 < a2), s2, (a1 < a2));

				gao.add_Edge(s1, (a1 < a2), a1, (s1 < s2));

				gao.add_Edge(a2, (s1 < s2), s2, (a1 < a2));

				gao.add_Edge(a2, (s1 < s2), a1, (s1 < s2));

			}

		}

		printf("%s\n", gao.solve() ? "Yes" : "No");

	}

	return 0;

}