POJ 3207 Ikki's Story IV - Panda's Trick

id=3207" target="_blank" style="">题目链接

题意：一个圆上顺序n个点，然后有m组连线，连接两点，要求这两点能够往圆内或圆外。问能否构造出使得满足全部线段不相交

思路：2-sat，推断相交的建边，一个在内。一个在外，然后跑一下2-sat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);

	}

	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, m, l[N], r[N];

int main() {

	while (~scanf("%d%d", &n, &m)) {

		gao.init(m);

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

			scanf("%d%d", &l[i], &r[i]);

			if (l[i] > r[i]) swap(l[i], r[i]);

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

				if ((l[i] > l[j] && l[i] < r[j] && r[j] > l[i] && r[j] < r[i]) || (r[i] > l[j] && r[i] < r[j] && l[j] > l[i] && r[j] < r[i])) {

					gao.add_Edge(i, 0, j, 0);

					gao.add_Edge(i, 1, j, 1);

				}

			}

		}

		printf("%s\n", gao.solve() ? "panda is telling the truth..." : "the evil panda is lying again");

	}

	return 0;

}