Bit Magic

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3209    Accepted Submission(s): 907


Problem Description Yesterday, my teacher taught me about bit operators: and (&), or (|), xor (^). I generated a number table a[N], and wrote a program to calculate the matrix table b[N][N] using three kinds of bit operator. I thought my achievement would get teacher's attention.
The key function is the code showed below.

There is no doubt that my teacher raised lots of interests in my work and was surprised to my talented programming skills. After deeply thinking, he came up with another problem: if we have the matrix table b[N][N] at first, can you check whether corresponding number table a[N] exists?  
Input There are multiple test cases.
For each test case, the first line contains an integer N, indicating the size of the matrix. (1 <= N <= 500).
The next N lines, each line contains N integers, the jth integer in ith line indicating the element b[i][j] of matrix. (0 <= b[i][j] <= 2 ³¹ - 1)
 
Output For each test case, output "YES" if corresponding number table a[N] exists; otherwise output "NO".  
Sample Input

2 0 4 4 0 3 0 1 24 1 0 86 24 86 0  
Sample Output

YES NO

列舉每一位，每次做一次2sat

#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <string>
#include <functional>
#include <cmath>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <stack>
using namespace std;
#define esp  1e-8
const double PI = acos(-1.0);
const double e = 2.718281828459;
const int inf = 2147483647;
const long long mod = 1000000007;
typedef long long ll;
void fre()
{
	freopen("alex.txt", "r", stdin);
	freopen("alex.txt", "w", stdout);
}
inline void in(int &x)
{
	register char ch;
	while (ch = getchar(), (ch < '0' || ch > '9'));
	x = ch - '0';
	while (ch = getchar(), ch >= '0' && ch <= '9') x = x * 10 + ch - '0';
}
inline void out(int x)
{
	register char hc[30];
	register int len = 0;
	hc[len++] = x % 10 + '0';
	while (x /= 10) hc[len++] = x % 10 + '0';
	for (int i = len - 1; i >= 0; i--) putchar(hc[i]);
}
const int maxn = 1015;
struct node
{
	int from, to, next;
}edge[4000006];
int hand[maxn], belong[maxn], dfn[maxn], low[maxn], vis[maxn], st[maxn];
int fp[maxn];//建立對應SCC編號的對映   
int color[maxn];//染色
int In[maxn];//新圖SCC的入度
vector<int>G[maxn];//縮點後新圖
int tot, num, index, cnt, n, m;
void addedge(int a, int b)
{
	edge[tot].from = a;
	edge[tot].to = b;
	edge[tot].next = hand[a];
	hand[a] = tot++;
}
void init()
{
	memset(hand, -1, sizeof(hand));
	memset(belong, -1, sizeof(belong));
	memset(dfn, 0, sizeof(dfn));
	memset(low, 0, sizeof(low));
	memset(vis, 0, sizeof(vis));
	memset(st, 0, sizeof(st));
	tot = 0;
	num = 0;
	index = 0;
	cnt = 0;
}
void Tarjan(int u)
{
	dfn[u] = low[u] = index++;
	vis[u] = 1;
	st[num++] = u;
	for (int i = hand[u]; i != -1; i = edge[i].next)
	{
		int v = edge[i].to;
		if (!dfn[v])
		{
			Tarjan(v);
			low[u] = min(low[u], low[v]);
		}
		else if (vis[v])
		{
			low[u] = min(low[u], dfn[v]);
		}
	}
	if (low[u] == dfn[u])
	{
		cnt++;
		int v;
		do
		{
			v = st[--num];
			vis[v] = 0;
			belong[v] = cnt;
		} while (u != v);
	}
}

int b[515][515];
int main()
{
	int i, j;
	
	while (~scanf("%d", &n))
	{
		int res = 0;
		for (i = 0; i < n; ++i)
		{
			for (j = 0; j < n; ++j)
			{
				scanf("%d", &b[i][j]);
				//in(b[i][j]);
			}
		}
		for (int k = 0; k < 31; ++k)
		{
			init();
			for (i = 0; i < n; ++i)
			{
				for (j = i + 1; j < n; ++j)
				{
					int p = (b[i][j] & (1 << k));
					if (i % 2 == 1 && j % 2 == 1)
					{
						if (p)
						{
							addedge(i, j + n); // ~a -> b
							addedge(j, i + n);// ~b -> a
						}
						else
						{
							addedge(i + n, i); // a -> ~a
							addedge(j + n, j); // b -> ~b
						}
					}
					else if (i % 2 == 0 && j % 2 == 0)
					{
						if (p)
						{
							addedge(i, i + n); // ~a -> a
							addedge(j, j + n); // ~b -> b
							addedge(i + n, j + n); // a -> b
							addedge(j + n, i + n); // b -> a
						}
						else
						{
							addedge(i + n, j); // a ->~b;
							addedge(j + n, i); // b -> ~a
						}
					}
					else
					{
						if (p)
						{
							addedge(i, j + n);//~a -> b;
							addedge(j, i + n);//~b -> a;
							addedge(i + n, j); // a -> ~b
							addedge(j + n, i); //b -> ~a
						}
						else
						{
							addedge(i, j);  //~a -> ~b;
							addedge(j, i); // ~b -> ~a;
							addedge(i + n, j + n); //a -> b
							addedge(j + n, i + n); // b -> a
						}
				
					}
				}
				
			}
			for (int i = 0; i < n * 2; ++i)
			{
				if (!dfn[i])
					Tarjan(i);

			}
			for (i = 0; i < n; ++i)
			{
				if (belong[i] == belong[i + n])
				{
					res = 1;
					break;
				}
			}
			if (res)
				break;
		}
		if (res)
			printf("NO\n");
		else
			printf("YES\n");
	}
}