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 = 20;
const int INF = 0x3f3f3f3f;
int a[MAXN][MAXN];
int b[MAXN][MAXN];
int ans[MAXN][MAXN];
int dx[5] = {-1, 0, 0, 0, 1};
int dy[5] = {0, 1, 0, -1, 0};
int n, m;

int get_col(int x, int y)   {
    int ret = a[x][y];
    for (int i=0; i<5; ++i) {
        int tx = x + dx[i], ty = y + dy[i];
        if (tx < 0 || tx >= n || ty < 0 || ty >= m) continue;
        ret += b[tx][ty];
    }
    return ret & 1;
}

int check(int s)    {
    memset (b, 0, sizeof (b));
    for (int i=0; i<m; ++i) {
        if (s & (1 << i))   b[0][i] = 1;
    }
    for (int i=1; i<n; ++i) {
        for (int j=0; j<m; ++j) {
            if (get_col (i-1, j))   b[i][j] = 1;
        }
    }
    for (int i=0; i<m; ++i) if (get_col (n-1, i))   return INF;
    int ret = 0;
    for (int i=0; i<n; ++i) {
        for (int j=0; j<m; ++j) ret += b[i][j];
    }
    return ret;
}

int main(void)  {       //POJ 3279 Fliptile
    while (scanf ("%d%d", &n, &m) == 2) {
        for (int i=0; i<n; ++i)    {
            for (int j=0; j<m; ++j)    {
                scanf ("%d", &a[i][j]);
            }
        }
        int mn = INF;
        memset (ans, 0, sizeof (ans));
        for (int i=0; i<(1<<m); ++i)    {
            int res = check (i);
            if (res < mn)   {
                mn = res;
                for (int i=0; i<n; ++i)    {
                    for (int j=0; j<m; ++j)    {
                        if (b[i][j])    ans[i][j] = 1;
                        else    ans[i][j] = 0;
                    }
                }
            }
        }
        if (mn < INF)   {
            for (int i=0; i<n; ++i)    {
                for (int j=0; j<m; ++j)    printf ("%d%c", ans[i][j], (j == m-1) ? '
' : ' ');
            }
        }
        else    puts ("IMPOSSIBLE");
    }

    return 0;
}