Fliptile POJ 3279（反转问题，开关问题）

Fliptile

Farmer John knows that an intellectually satisfied cow is a happy cow who will give more milk. He has arranged a brainy activity for cows in which they manipulate an M × N grid (1 ≤ M ≤ 15; 1 ≤ N ≤ 15) of square tiles, each of which is colored black on one side and white on the other side.

As one would guess, when a single white tile is flipped, it changes to black; when a single black tile is flipped, it changes to white. The cows are rewarded when they flip the tiles so that each tile has the white side face up. However, the cows have rather large hooves and when they try to flip a certain tile, they also flip all the adjacent tiles (tiles that share a full edge with the flipped tile). Since the flips are tiring, the cows want to minimize the number of flips they have to make.

Help the cows determine the minimum number of flips required, and the locations to flip to achieve that minimum. If there are multiple ways to achieve the task with the minimum amount of flips, return the one with the least lexicographical ordering in the output when considered as a string. If the task is impossible, print one line with the word "IMPOSSIBLE".

Input
Line 1: Two space-separated integers: M and N
Lines 2.. M+1: Line i+1 describes the colors (left to right) of row i of the grid with N space-separated integers which are 1 for black and 0 for white
Output
Lines 1.. M: Each line contains N space-separated integers, each specifying how many times to flip that particular location.
Sample Input

4 4
1 0 0 1
0 1 1 0
0 1 1 0
1 0 0 1

题意：给定一个MxN的格子，每个格子可以反转正反面，一面黑色1表示，一面是白色0表示，规则是，反转一个格子，和它相邻的格子都会被反转，目标是把所有格子全都反转成白色，求最小反转次数，如果有多种方案输出字典序最小的一组，解不存在输出IMPOSSIBLE

思路：首先同一个格子反转两次就会恢复原状，所以多次反转是多余的，此外，反转的格子的集合相同的话，其次序是无关紧要的。
因此总共有 2NM 种翻转方法。不过解的空间太大了，需要更有效的方法。
让我们回顾下前面的问题前面的问题传送门，那道题中，让最左端的牛反转的方法只有一种，于是直接判断的方法确定就可以了。
不妨先看看样例中最左上角的格子，在这里除了翻转(1,1)之外，翻转(1,2),(2,1)也可以把这个格子翻转，所以之前的方法行不通。
所以不妨先指定好最上面一行的翻转方法，此时能够翻转(1,1)的就只剩下(2,1),所以可以直接判断(2,1)是否需要翻转，类似的(2,1)~(2,N)都能这样判断，如此反复下去就可以确定所有格子的翻转方法，最后(M,1)~(M,N)如果并非全是白色，就说明不存在可行方法
这样先确定第一行的翻转方式，然后可以容易判断这样是否存在解以及解的最小步数是多少，这样将第一行的所有的反转方式都枚举尝试一遍就能求出整个问题的最小步数，复杂度 O（MN2N）
code：

#include 
#include 
#include 
using namespace std;
const int MAXN = 20;
const int dx[5] = {-1,0,0,0,1};
const int dy[5] = {0,-1,0,1,0};
int M,N;
int tile[MAXN][MAXN];
int opt[MAXN][MAXN]; //保存最优解
int flip[MAXN][MAXN]; //保存中间结果1代表反转0不变

int get(int x,int y){//四个方向的反转和，如果是的当前位置和为奇数说明黑色，否则白色
    int c = tile[x][y];
    for(int d = 0; d < 5; d++){
        int xx = x + dx[d];
        int yy = y + dy[d];
        if(xx >= 0 && xx < M && yy >= 0 && yy < N)
            c += flip[xx][yy];
    }
    return c % 2;
}

int calc(){
    //求出从第2行开始的反转方法
    for(int i = 1; i < M; i++){
        for(int j = 0; j < N; j++){
            if(get(i-1,j) != 0){
                // (i-1,j)是黑色的话，则必须反转这个格子
                flip[i][j] = 1;
            }
        }
    }
    //判断最后一行是否全白
    for(int j = 0; j < N; j++){
        if(get(M-1,j) != 0) //无解
            return -1;
    }
    int res = 0;
    for(int i = 0; i < M; i++){
        for(int j = 0; j < N; j++){
            res += flip[i][j];
        }
    }
    return res;
}

void solve(){
    int res = -1;
    //按照字典序尝试第一行所有可能性
    for(int i = 0; i < 1 << N; i++){
        memset(flip,0,sizeof(flip));
        for(int j = 0; j < N; j++){
            flip[0][N-j-1] = i >> j & 1;
        }
        int num = calc();
        if(num > 0 && (res < 0 || res > num)){
            res = num;
            memcpy(opt,flip,sizeof(flip));
        }
    }
    if(res < 0) //无解
        printf("IMPOSSIBLE\n");
    else{
        for(int i = 0; i < M; i++){
            for(int j = 0; j < N; j++){
                printf("%d%c",opt[i][j],j + 1 == N ? '\n' : ' ');
            }
        }
    }
}
int main(){
    scanf("%d%d",&M,&N);
    for(int i = 0; i < M; i++){
        for(int j = 0; j < N; j++){
            scanf("%d",&tile[i][j]);
        }
    }
    solve();
    return 0;
}

from 挑战程序设计竞赛