CodeForces 486B OR in Matrix

题目链接：http://codeforces.com/contest/486/problem/B

题目大意：

　　有两个矩阵A和B，矩阵B的元素B_ij的值是所有A的第i行元素和第j列元素或运算后的值。现在给出B矩阵，求A矩阵。

分析：

　　首先，如果B[i][j] == 1，那么B矩阵或者第i行全为1，或者第j列全为1；如果B[i][j] == 0，那么A矩阵第i行全为0且第j列也全为0。

　　其次，如果A[i][j] == 1，那么B矩阵第i行全为1且第j列也全为1。以此可以推出B矩阵全为1的行和全为1的列要么全有，要么全没有。

　　于是可以把B矩阵全为1的行和全为1的列找出来一一配对，交点位置对应的A矩阵位置置1。

代码如下：

#include <bits/stdc++.h>
using namespace std;
 
#define rep(i,n) for (int i = 0; i < (n); ++i)
#define For(i,s,t) for (int i = (s); i <= (t); ++i)
#define rFor(i,t,s) for (int i = (t); i >= (s); --i)
#define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
#define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)
 
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
 
#define LOWBIT(x) ((x)&(-x))
 
#define ALL(x) x.begin(),x.end()
#define INS(x) inserter(x,x.begin())
 
#define ms0(a) memset(a,0,sizeof(a))
#define msI(a) memset(a,inf,sizeof(a))
#define msM(a) memset(a,-1,sizeof(a))
 
#define pii pair<int,int> 
#define piii pair<pair<int,int>,int> 
#define mp make_pair
#define pb push_back
#define fi first
#define se second
 
inline int gc(){
    static const int BUF = 1e7;
    static char buf[BUF], *bg = buf + BUF, *ed = bg;
     
    if(bg == ed) fread(bg = buf, 1, BUF, stdin);
    return *bg++;
} 
 
inline int ri(){
    int x = 0, f = 1, c = gc();
    for(; c<48||c>57; f = c=='-'?-1:f, c=gc());
    for(; c>47&&c<58; x = x*10 + c - 48, c=gc());
    return x*f;
}
 
typedef long long LL;
typedef unsigned long long uLL;
const int inf = 1e9 + 9;
const LL mod = 1e9 + 7;
const int maxN = 1e2 + 7;
 
int n, m;
int A[maxN][maxN];
// Brow[i]表示第i行所有数与运算的值 
// Bcol[i]表示第i列所有数与运算的值 
// AllBrow表示所有Brow[i]或运算的值 
// AllBcol表示所有Bcol[i]或运算的值 
int B[maxN][maxN], Brow[maxN], AllBrow, Bcol[maxN], AllBcol;
bool ans;
 
int main(){
    while(cin >> n >> m) {
        ms0(A);
        rep(i, n) rep(j, m) cin >> B[i][j];
         
        AllBrow = AllBcol = 0; 
        rep(i, n) {
            Brow[i] = 1;
            rep(j, m) Brow[i] &= B[i][j];
            AllBrow |= Brow[i];
        }
        rep(i, m) {
            Bcol[i] = 1;
            rep(j, n) Bcol[i] &= B[j][i];
            AllBcol |= Bcol[i];
        }
         
        ans = !(AllBrow ^ AllBcol);
         
        rep(i, n) rep(j, m) {
            if(B[i][j] == 1 && Brow[i] == 0 && Bcol[j] == 0) {
                ans = false;
                i = n;
                break;
            }
        }
         
        if(ans) {
            rep(i, n) rep(j, m) {
                if(Brow[i] & Bcol[j]) {
                    A[i][j] = 1;
                }
            }
             
            cout << "YES" << endl;
            rep(i, n) {
                rep(j, m) {
                    cout << A[i][j] << " ";
                }
                cout << endl;
            }
        }
        else cout << "NO" << endl;
    }
    return 0;
}