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

 template<class T>T gcd(T a, T b){return b==?a:gcd(b,a%b);}
 template<class T>void ReadInt(T &x){char c=getchar();while(!isdigit(c))c=getchar();x=;while(isdigit(c)){x=x*+c-'';c=getchar();}}
 template<class T>void WriteInt(T i) {int p=;static int b[];if(i == ) b[p++] = ;else while(i){b[p++]=i%;i/=;}for(int j=p-;j>=;j--)pchr(''+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 >=  || m >= ) {
         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 = ;
         copy_rec2();
         rep_up0(i, m) {
             copy_rec();
             swap_col(, i);
             copy_rec2();
             int flag[];
             rep_up0(j, n) {
                 flag[j] = a[j][] != b[j][];
             }
             rep_up0(j, n) {
                 for (int k = ; k < m; k++) {
                     a[j][k] ^= flag[j];
                 }
             }
             int yes = ;
             for (int j = ; j < m; j++) {
                 int ok0 = ;
                 for (int k = j; k < m; k++) {
                     if (equal_col(k, j)) {
                         swap_col(k, j);
                         ok0 = ;
                         break;
                     }
                 }
                 if (!ok0) {
                     yes = ;
                     break;
                 }
             }
             if (yes) {
                 ok = ;
                 break;
             }
         }
         puts(ok? "Yes" : "No");
     }
     return ;
 }