HDU 4421 Bit Magic(2-sat）

HDU 4421 Bit Magic

题目链接

题意：就依据题目，给定b数组。看能不能构造出一个符合的a数组

思路：把每一个数字的每一个二进制位单独考虑。就变成一个2-sat题目了，依据题目中的式子建立2-sat的边。然后每一位跑2-sat。假设每位都符合。就是YES，假设有一位不符合就是NO

代码：

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

const int MAXNODE = 505;

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, b[N][N];

bool solve() {
	for (int k = 0; k < 31; k++) {
		gao.init(n);
		for (int i = 0; i < n; i++)
			for (int j = 0; j < n; j++) {
				int tmp = (b[i][j]>>k)&1;
				if (i == j) {
					if (tmp) return false;
				}
				else if (i % 2 == 1 && j % 2 == 1) {
					if (tmp) gao.add_Edge(i, tmp, j, tmp);
					else {
						gao.add_Edge(i, tmp, i, tmp);
						gao.add_Edge(j, tmp, j, tmp);
					}
				}
				else if (i % 2 == 0 && j % 2 == 0) {
					if (tmp) {
						gao.add_Edge(i, tmp, i, tmp);
						gao.add_Edge(j, tmp, j, tmp);
					} else gao.add_Edge(i, tmp, j, tmp);
				} else {
					if (tmp) {
						gao.add_Edge(i, tmp, j, tmp);
						gao.add_Edge(i, !tmp, j, !tmp);
					} else {
						gao.add_Edge(i, tmp, j, !tmp);
						gao.add_Edge(i, !tmp, j, tmp);
					}
				}
			}
		if (!gao.solve()) return false;
	}
	return true;
}

int main() {
	while (~scanf("%d", &n)) {
		for (int i = 0; i < n; i++)
			for (int j = 0; j < n; j++)
				scanf("%d", &b[i][j]);
		printf("%s
", solve() ? "YES" : "NO");
	}
	return 0;
}