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\n", solve() ? "YES" : "NO");}return 0;
}