[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：

  1 #pragma comment(linker, "/STACK:10240000,10240000")
  2
  3 #include <iostream>
  4 #include <cstdio>
  5 #include <algorithm>
  6 #include <cstdlib>
  7 #include <cstring>
  8 #include <map>
  9 #include <queue>
 10 #include <deque>
 11 #include <cmath>
 12 #include <vector>
 13 #include <ctime>
 14 #include <cctype>
 15 #include <set>
 16 #include <bitset>
 17 #include <functional>
 18 #include <numeric>
 19 #include <stdexcept>
 20 #include <utility>
 21
 22 using namespace std;
 23
 24 #define mem0(a) memset(a, 0, sizeof(a))
 25 #define lson l, m, rt << 1
 26 #define rson m + 1, r, rt << 1 | 1
 27 #define define_m int m = (l + r) >> 1
 28 #define rep_up0(a, b) for (int a = 0; a < (b); a++)
 29 #define rep_up1(a, b) for (int a = 1; a <= (b); a++)
 30 #define rep_down0(a, b) for (int a = b - 1; a >= 0; a--)
 31 #define rep_down1(a, b) for (int a = b; a > 0; a--)
 32 #define all(a) (a).begin(), (a).end()
 33 #define lowbit(x) ((x) & (-(x)))
 34 #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) {}
 35 #define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {}
 36 #define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {}
 37 #define pchr(a) putchar(a)
 38 #define pstr(a) printf("%s", a)
 39 #define sstr(a) scanf("%s", a);
 40 #define sint(a) ReadInt(a)
 41 #define sint2(a, b) ReadInt(a);ReadInt(b)
 42 #define sint3(a, b, c) ReadInt(a);ReadInt(b);ReadInt(c)
 43 #define pint(a) WriteInt(a)
 44 #define if_else(a, b, c) if (a) { b; } else { c; }
 45 #define if_than(a, b) if (a) { b; }
 46 #define test_print1(a) cout << "var1 = " << a << endl
 47 #define test_print2(a, b) cout << "var1 = " << a << ", var2 = " << b << endl
 48 #define test_print3(a, b, c) cout << "var1 = " << a << ", var2 = b" << ", var3 = " << c << endl
 49
 50 typedef double db;
 51 typedef long long LL;
 52 typedef pair<int, int> pii;
 53 typedef multiset<int> msi;
 54 typedef set<int> si;
 55 typedef vector<int> vi;
 56 typedef map<int, int> mii;
 57
 58 const int dx[8] = {0, 0, -1, 1};
 59 const int dy[8] = {-1, 1, 0, 0};
 60 const int maxn = 1e2 + 7;
 61 const int maxm = 1e3 + 7;
 62 const int maxv = 1e7 + 7;
 63 const int max_val = 1e6 + 7;
 64 const int MD = 22;
 65 const int INF = 1e9 + 7;
 66 const double pi = acos(-1.0);
 67 const double eps = 1e-10;
 68
 69 template<class T>T gcd(T a, T b){return b==0?a:gcd(b,a%b);}
 70 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();}}
 71 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]);}
 72 template<class T>bool max_update(T &a,const T &b){if(b>a){a = b; return true;}return false;}
 73 template<class T>bool min_update(T &a,const T &b){if(b<a){a = b; return true;}return false;}
 74 template<class T>T condition(bool f, T a, T b){return f?a:b;}
 75 template<class T>void copy_arr(T a[], T b[], int n){rep_up0(i,n)a[i]=b[i];}
 76 int make_id(int x, int y, int n) { return x * n + y; }
 77
 78 int n, m;
 79 int a[maxn][maxn], b[maxn][maxn], buf[maxn][maxn];
 80
 81 int swap_col(int y1, int y2) {
 82     rep_up0(i, n) {
 83         swap(a[i][y1], a[i][y2]);
 84     }
 85 }
 86
 87 int equal_col(int y1, int y2) {
 88     rep_up0(i, n) {
 89         if (a[i][y1] != b[i][y2]) return false;
 90     }
 91     return true;
 92 }
 93
 94 int copy_rec() {
 95     rep_up0(i, n) {
 96         rep_up0(j, m) {
 97             a[i][j] = buf[i][j];
 98         }
 99     }
100 }
101 int copy_rec2() {
102     rep_up0(i, n) {
103         rep_up0(j, m) {
104             buf[i][j] = a[i][j];
105         }
106     }
107 }
108
109 int main() {
110     //freopen("in.txt", "r", stdin);
111     //freopen("out.txt", "w", stdout);
112     while (cin >> n >> m, n >= 0 || m >= 0) {
113         rep_up0(i, n) {
114             rep_up0(j, m) {
115                 sint(a[i][j]);
116             }
117         }
118         rep_up0(i, n) {
119             rep_up0(j, m) {
120                 sint(b[i][j]);
121             }
122         }
123         int ok = 0;
124         copy_rec2();
125         rep_up0(i, m) {
126             copy_rec();
127             swap_col(0, i);
128             copy_rec2();
129             int flag[100];
130             rep_up0(j, n) {
131                 flag[j] = a[j][0] != b[j][0];
132             }
133             rep_up0(j, n) {
134                 for (int k = 1; k < m; k++) {
135                     a[j][k] ^= flag[j];
136                 }
137             }
138             int yes = 1;
139             for (int j = 1; j < m; j++) {
140                 int ok0 = 0;
141                 for (int k = j; k < m; k++) {
142                     if (equal_col(k, j)) {
143                         swap_col(k, j);
144                         ok0 = 1;
145                         break;
146                     }
147                 }
148                 if (!ok0) {
149                     yes = 0;
150                     break;
151                 }
152             }
153             if (yes) {
154                 ok = 1;
155                 break;
156             }
157         }
158         puts(ok? "Yes" : "No");
159     }
160     return 0;
161 }