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

HDU 1816， POJ 2723 Get Luffy Out

pid=1816" target="_blank" style="">题目链接

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;
}