[hdu3484]枚举

题意：给两个个01矩阵，有两种操作，（1）交换两列（2）反转某一行。求能否通过若干操作使两矩阵相等

思路：（把所有对B的操作放到A上来，这一定是可以做到一样的效果的）枚举B矩阵的第一列对应A矩阵的第几列，交换这两列，那么根据两个矩阵这一列的相等情况可以确定每一行的操作情况（操作次数实际上只有0和1两种情况），然后根据这个情况确定执行了所有行操作的A矩阵，然后从第二列开始到第m列，依次用A矩阵的某一列（这个也是枚举）去“匹配”B矩阵的当前列，匹配好了那么继续后面的匹配（任何一个可以匹配当前列的列他们是没有区别的，任取一个，所以我们取第一次出现的那一个）。这样操作直到完全匹配好。详见代码：

#pragma comment(linker, "/STACK:10240000,10240000")

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <map>
#include <queue>
#include <deque>
#include <cmath>
#include <vector>
#include <ctime>
#include <cctype>
#include <set>
#include <bitset>
#include <functional>
#include <numeric>
#include <stdexcept>
#include <utility>

using namespace std;

#define mem0(a) memset(a, 0, sizeof(a))
#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1
#define define_m int m = (l + r) >> 1
#define rep_up0(a, b) for (int a = 0; a < (b); a++)
#define rep_up1(a, b) for (int a = 1; a <= (b); a++)
#define rep_down0(a, b) for (int a = b - 1; a >= 0; a--)
#define rep_down1(a, b) for (int a = b; a > 0; a--)
#define all(a) (a).begin(), (a).end()
#define lowbit(x) ((x) & (-(x)))
#define constructInt4(name, a, b, c, d) name(int a = 0, int b = 0, int c = 0, int d = 0): a(a), b(b), c(c), d(d) {}
#define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {}
#define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {}
#define pchr(a) putchar(a)
#define pstr(a) printf("%s", a)
#define sstr(a) scanf("%s", a);
#define sint(a) ReadInt(a)
#define sint2(a, b) ReadInt(a);ReadInt(b)
#define sint3(a, b, c) ReadInt(a);ReadInt(b);ReadInt(c)
#define pint(a) WriteInt(a)
#define if_else(a, b, c) if (a) { b; } else { c; }
#define if_than(a, b) if (a) { b; }
#define test_print1(a) cout << "var1 = " << a << endl
#define test_print2(a, b) cout << "var1 = " << a << ", var2 = " << b << endl
#define test_print3(a, b, c) cout << "var1 = " << a << ", var2 = b" << ", var3 = " << c << endl

typedef double db;
typedef long long LL;
typedef pair<int, int> pii;
typedef multiset<int> msi;
typedef set<int> si;
typedef vector<int> vi;
typedef map<int, int> mii;

const int dx[8] = {0, 0, -1, 1};
const int dy[8] = {-1, 1, 0, 0};
const int maxn = 1e2 + 7;
const int maxm = 1e3 + 7;
const int maxv = 1e7 + 7;
const int max_val = 1e6 + 7;
const int MD = 22;
const int INF = 1e9 + 7;
const double pi = acos(-1.0);
const double eps = 1e-10;

template<class T>T gcd(T a, T b){return b==0?a:gcd(b,a%b);}
template<class T>void ReadInt(T &x){char c=getchar();while(!isdigit(c))c=getchar();x=0;while(isdigit(c)){x=x*10+c-'0';c=getchar();}}
template<class T>void WriteInt(T i) {int p=0;static int b[20];if(i == 0) b[p++] = 0;else while(i){b[p++]=i%10;i/=10;}for(int j=p-1;j>=0;j--)pchr('0'+b[j]);}
template<class T>bool max_update(T &a,const T &b){if(b>a){a = b; return true;}return false;}
template<class T>bool min_update(T &a,const T &b){if(b<a){a = b; return true;}return false;}
template<class T>T condition(bool f, T a, T b){return f?a:b;}
template<class T>void copy_arr(T a[], T b[], int n){rep_up0(i,n)a[i]=b[i];}
int make_id(int x, int y, int n) { return x * n + y; }

int n, m;
int a[maxn][maxn], b[maxn][maxn], buf[maxn][maxn];

int swap_col(int y1, int y2) {
    rep_up0(i, n) {
        swap(a[i][y1], a[i][y2]);
    }
}

int equal_col(int y1, int y2) {
    rep_up0(i, n) {
        if (a[i][y1] != b[i][y2]) return false;
    }
    return true;
}

int copy_rec() {
    rep_up0(i, n) {
        rep_up0(j, m) {
            a[i][j] = buf[i][j];
        }
    }
}
int copy_rec2() {
    rep_up0(i, n) {
        rep_up0(j, m) {
            buf[i][j] = a[i][j];
        }
    }
}

int main() {
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    while (cin >> n >> m, n >= 0 || m >= 0) {
        rep_up0(i, n) {
            rep_up0(j, m) {
                sint(a[i][j]);
            }
        }
        rep_up0(i, n) {
            rep_up0(j, m) {
                sint(b[i][j]);
            }
        }
        int ok = 0;
        copy_rec2();
        rep_up0(i, m) {
            copy_rec();
            swap_col(0, i);
            copy_rec2();
            int flag[100];
            rep_up0(j, n) {
                flag[j] = a[j][0] != b[j][0];
            }
            rep_up0(j, n) {
                for (int k = 1; k < m; k++) {
                    a[j][k] ^= flag[j];
                }
            }
            int yes = 1;
            for (int j = 1; j < m; j++) {
                int ok0 = 0;
                for (int k = j; k < m; k++) {
                    if (equal_col(k, j)) {
                        swap_col(k, j);
                        ok0 = 1;
                        break;
                    }
                }
                if (!ok0) {
                    yes = 0;
                    break;
                }
            }
            if (yes) {
                ok = 1;
                break;
            }
        }
        puts(ok? "Yes" : "No");
    }
    return 0;
}