状态压缩+枚举 POJ 3279 Fliptile

题目传送门

 /*
     题意：问最少翻转几次使得棋子都变白，输出翻转的位置
     状态压缩+枚举：和之前UVA_11464差不多，枚举第一行，可以从上一行的状态知道当前是否必须翻转
 */
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 using namespace std;

 const int MAXN = ;
 const int INF = 0x3f3f3f3f;
 int a[MAXN][MAXN];
 int b[MAXN][MAXN];
 int ans[MAXN][MAXN];
 int dx[] = {-, , , , };
 int dy[] = {, , , -, };
 int n, m;

 int get_col(int x, int y)   {
     int ret = a[x][y];
     for (int i=; i<; ++i) {
         int tx = x + dx[i], ty = y + dy[i];
         if (tx <  || tx >= n || ty <  || ty >= m) continue;
         ret += b[tx][ty];
     }
     return ret & ;
 }

 int check(int s)    {
     memset (b, , sizeof (b));
     for (int i=; i<m; ++i) {
         if (s & ( << i))   b[][i] = ;
     }
     for (int i=; i<n; ++i) {
         for (int j=; j<m; ++j) {
             if (get_col (i-, j))   b[i][j] = ;
         }
     }
     for (int i=; i<m; ++i) if (get_col (n-, i))   return INF;
     int ret = ;
     for (int i=; i<n; ++i) {
         for (int j=; j<m; ++j) ret += b[i][j];
     }
     return ret;
 }

 int main(void)  {       //POJ 3279 Fliptile
     while (scanf ("%d%d", &n, &m) == ) {
         for (int i=; i<n; ++i)    {
             for (int j=; j<m; ++j)    {
                 scanf ("%d", &a[i][j]);
             }
         }
         int mn = INF;
         memset (ans, , sizeof (ans));
         for (int i=; i<(<<m); ++i)    {
             int res = check (i);
             if (res < mn)   {
                 mn = res;
                 for (int i=; i<n; ++i)    {
                     for (int j=; j<m; ++j)    {
                         if (b[i][j])    ans[i][j] = ;
                         else    ans[i][j] = ;
                     }
                 }
             }
         }
         if (mn < INF)   {
             for (int i=; i<n; ++i)    {
                 for (int j=; j<m; ++j)    printf ("%d%c", ans[i][j], (j == m-) ? '\n' : ' ');
             }
         }
         else    puts ("IMPOSSIBLE");
     }

     return ;
 }