POJ 3279 Fliptile

//这道题还是很有意思
//最开始无从下手但是 可以发现 翻牌和先后顺序无关 ---> 那么只能翻1次或不翻 因为其他情况和1 0 都是重复的
//那么 如果强行枚举就是2^n
//这里就没有想到了 ------->降维度的简化
//确定第一行后 第二行就要把第一行的1都翻为0 才能继续下去 所以第二行事实是确定的
//以此类推 如果最后一行都为0则成立否则 失败
//那么只需要枚举第一行的情况 1e4的复杂度 很小

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <queue>
#include <string>
#include <algorithm>
#include <set>
#define READ() freopen("in.txt", "r", stdin);
#define WRITE() freopen("out.txt", "w", stdout);
#define INF 0x3f3f3f3f

using namespace std;

int M, N;
int maze[28][28], mazecp[28][28];
int perm[28], num = 0;
set<string> s;
int d[4][2] =
{
    -1, 0,
    0, 1,
    1, 0,
    0, -1
};
char operation[28][28];
char ans_operation[28][28];
int tmp_ans = 0, ans = INF;

bool OK(int x, int y)
{
    if (x < 0|| x >= M || y < 0 || y >= N) return false;
    return true;
}


void dfs(int x)//x表示现在在第几列
{
    if (x == N)//最后一列已经决定翻完
    {
        //先把第一行的翻完
        tmp_ans = 0;
        for (int i = 0; i < N; i++)
        {
            if (operation[0][i] == '1')
            {
                tmp_ans++;
                mazecp[0][i] += 1;
                for (int p = 0; p < 4; p++)
                {
                    int nx = 0+d[p][0], ny = i+d[p][1];
                    if (OK(nx, ny))
                    {
                        mazecp[nx][ny] += 1;
                    }
                }
            }
        }
        //第一行确定 其余所有行都确定
        for (int i = 1; i < M; i++)
        {
            for (int j = 0; j < N; j++)
            {
                if (mazecp[i-1][j] % 2 == 1)
                {
                    operation[i][j] = '1';//那么这一个就要翻
                    tmp_ans++;
                    mazecp[i][j] += 1;
                    for (int p = 0; p < 4; p++)//造成的影响
                    {
                        int nx = i+d[p][0], ny = j+d[p][1];
                        if (OK(nx, ny))
                        {
                            mazecp[nx][ny] += 1;
                        }
                    }
                }
            }
        }
        //检查最后一行是否为 都为偶数
        bool success = true;
        for (int i = 0; i < N; i++)
        {
            if (mazecp[M-1][i] % 2 == 1)
            {
                success = false;
                break;
            }
        }
        if (success)
        {
            if (tmp_ans < ans)//要先保证翻的次数最少 然后是最小字典序 哎 坑了
            {
                ans = tmp_ans;
                for (int i = 0; i < M; i++)
                {
                    for (int j = 0; j < N; j++) ans_operation[i][j] = operation[i][j];
                }
            }
            else if (tmp_ans == ans)
            {
                if (strcmp(ans_operation[0], operation[0]) > 0)
                {
                    for (int i = 0; i < M; i++)
                    for (int j = 0; j < N; j++) ans_operation[i][j] = operation[i][j];
                }
            }//成功
        }
        for (int i = 1; i < M; i++) fill(operation[i], operation[i]+N, '0');//重置操作
        for (int i = 0; i < M; i++)
        {
            for (int j = 0; j < N; j++) mazecp[i][j] = maze[i][j];//重置地图
        }
        return ;
    }
    operation[0][x] = '0';
    dfs(x+1);
    operation[0][x] = '1';
    dfs(x+1);
    return ;
}

int main()
{
    //READ()
    //WRITE()
    while(~scanf("%d%d", &M, &N))
    {
        for (int i = 0; i < M; i++)
        {
            for (int j = 0; j < N; j++)
            {
                scanf("%d", &maze[i][j]);
                mazecp[i][j] = maze[i][j];
            }
        }
        memset(operation, '0', sizeof(operation));
        memset(ans_operation, '1', sizeof(ans_operation));
        ans = INF;
        dfs(0);
        if (ans < INF)
        {
            for (int i = 0; i < M; i++)
            {
                for (int j = 0; j < N; j++)
                {
                    printf("%c ", ans_operation[i][j]);
                }
                putchar('\n');
            }
        }
        else printf("IMPOSSIBLE\n");
    }
    return 0;
}