HDU 1816， POJ 2723 Get Luffy Out（2-sat)

HDU 1816， POJ 2723 Get Luffy Out

题意：N串钥匙。每串2把，仅仅能选一把。然后有n个大门，每一个门有两个锁，开了一个就能通过，问选一些钥匙，最多能通过多少个门

思路：二分通过个数。然后对于钥匙建边至少一个不选，门建边至少一个选，然后2-sat搞一下就可以。

一開始是按每串钥匙为1个结点，但是后面发现数据有可能一把钥匙，出如今不同串（真是不合理），所以这个做法就跪了

代码：

#include <cstdio>

#include <cstring>

#include <cstdlib>

#include <vector>

#include <algorithm>

using namespace std;

const int MAXNODE = 2105;

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 = 2055;

int n, m;

int x[N], y[N];

int k1[N], k2[N];

bool judge(int d) {

	gao.init(2 * n);

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

		gao.add_Edge(x[i], 0, y[i], 0);

	for (int i = 1; i <= d; i++)

		gao.add_Edge(k1[i], 1, k2[i], 1);

	return gao.solve();

}

int main() {

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

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

			scanf("%d%d", &x[i], &y[i]);

		for (int i = 1; i <= m; i++)

			scanf("%d%d", &k1[i], &k2[i]);

		int l = 0, r = m + 1;

		while (l < r) {

			int mid = (l + r) / 2;

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

			else r = mid;

		}

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

	}

	return 0;

}