HDU 3715 Go Deeper

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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