hdu4421&zoj3656 Bit Magic(2SAT)

Bit Magic

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 649    Accepted Submission(s): 185

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
   
   
   
   

 

Source

2012 Asia ChangChun Regional Contest

这是12年长春现场赛的B，记忆非常深刻，当年还是太年轻了，没学2SAT竟然就去长春了。。。

这是一道明显的2SAT，由于位操作每一位是单独作用的，所以可以将31位的整数拆成31位，每一位跑一次2SAT判断是否合法，具体建图和这题是一样的，非常经典的建图。

不过这题还是跪了几个小时，检查了半天，原来是一个优先级搞错了。。。。

详情请见代码：

#include <iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int N = 505;
const int M = 1000005;
int table[N][N];
int head[N<<1];
int scc[N<<1];
int vis[N<<1];
int stack1[N<<1];
int stack2[N<<1];
struct edge
{
    int to,next;
}g[M];
int n,num;
bool flag;

void init()
{
    memset(head,-1,sizeof(head));
    memset(vis,0,sizeof(vis));
    memset(scc,0,sizeof(scc));
    flag = true;
    stack1[0] = stack2[0] = num = 0;
}

void build(int s,int e)
{
    g[num].to = e;
    g[num].next = head[s];
    head[s] = num ++;
}

void dfs(int cur,int &sig,int &cnt)
{
    if(!flag)
        return;
    vis[cur] = ++sig;
    stack1[++stack1[0]] = cur;
    stack2[++stack2[0]] = cur;
    for(int i = head[cur];i != -1;i = g[i].next)
    {
        if(!vis[g[i].to])
            dfs(g[i].to,sig,cnt);
        else
        {
            if(!scc[g[i].to])
            {
                while(vis[stack2[stack2[0]]] > vis[g[i].to])
                    stack2[0] --;
            }
        }
    }
    if(stack2[stack2[0]] == cur)
    {
        stack2[0] --;
        cnt ++;
        do
        {
            scc[stack1[stack1[0]]] = cnt;
            int tmp = stack1[stack1[0]];
            if((tmp >= n && scc[tmp - n] == cnt) || (tmp < n && scc[tmp + n] == cnt))
            {
                flag = false;
                return;
            }
        }while(stack1[stack1[0] --] != cur);
    }
}

void Gabow()
{
    int i,sig,cnt;
    sig = cnt = 0;
    for(i = 0;i < n + n && flag;i ++)
        if(!vis[i])
            dfs(i,sig,cnt);
}

int main()
{
    int i,j,k;
    while(scanf("%d",&n) != EOF)
    {
        for(i = 0;i < n;i ++)
            for(j = 0;j < n;j ++)
                scanf("%d",&table[i][j]);
        flag = true;
        for(i = 0;i < n;i ++)/*zoj要这一段才能过*/
        {
            if(table[i][i])
                flag = false;
        }
        if(flag == false)
        {
            printf("NO\n");
            continue;
        }
        for(k = 0;k < 32;k ++)
        {
            init();
            for(i = 0;i < n;i ++)
            {
                for(j = 0;j < n;j ++)
                {
                    if(i == j)
                        continue;
                    else
                    {
                        int ti = table[i][j] & (1<<k);
                        if((i&1) && (j&1))//or
                        {
                            if(ti)
                            {
                                build(i,j + n);
                                build(j,i + n);
                            }
                            else
                            {
                                build(i + n,i);
                                build(j + n,j);
                            }
                        }
                        else
                        {
                            if((i&1) == 0 && (j&1) == 0)//and   !!!!!!!!!!
                            {//==的优先级高于&！！！！！！！！！！！！！！
                                if(ti)
                                {
                                    build(i,i + n);
                                    build(j,j + n);
                                }
                                else
                                {
                                    build(i + n,j);
                                    build(j + n,i);
                                }
                            }
                            else//xor
                            {
                                if(ti)
                                {
                                    build(i,j + n);
                                    build(i + n,j);
                                    build(j,i + n);
                                    build(j + n,i);
                                }
                                else
                                {
                                    build(i,j);
                                    build(j,i);
                                    build(i + n,j + n);
                                    build(j + n,i + n);
                                }
                            }
                        }
                    }
                }
            }
            Gabow();
            if(flag == false)
                break;
        }
        if(flag)
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}
//1234MS	7156K