HDU 3715 Go Deeper（2-sat）

HDU 3715 Go Deeper

题目链接

题意：依据题意那个函数，构造x数组。问最大能递归层数

思路：转化为2-sat问题，因为x仅仅能是0。1，c仅仅能是0，1。2那么问题就好办了，对于0， 1， 2相应各自是3种表达式，然后二分深度，搞2-sat就可以

代码：

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <vector>
#include <algorithm>
using namespace std;

const int MAXNODE = 205;

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 = 10005;

int t, n, m;
int a[N], b[N], c[N];

bool judge(int dep) {
	gao.init(n);
	for (int i = 0; i < dep; i++) {
		if (c[i] == 0)
			gao.add_Edge(a[i], 1, b[i], 1);
		else if (c[i] == 1) {
			gao.add_Edge(a[i], 0, a[i], 1);
			gao.add_Edge(a[i], 0, b[i], 1);
			gao.add_Edge(b[i], 0, a[i], 1);
			gao.add_Edge(b[i], 0, b[i], 1);
		} else
			gao.add_Edge(a[i], 0, b[i], 0);
	}
	return gao.solve();
}

int main() {
	scanf("%d", &t);
	while (t--) {
		scanf("%d%d", &n, &m);
		for (int i = 0; i < m; i++)
			scanf("%d%d%d", &a[i], &b[i], &c[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;
}

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