约翰又来折腾牛了……

Description

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
Sample Output

0 0 0 0
1 0 0 1
1 0 0 1
0 0 0 0

题目是一道简单的开关问题（插句题外话，记得小时候玩过肯德基制作的一款Flash小游戏，就是根据开关问题来设计的。），要将一个M X N的黑白色相间的格子翻转为白色（翻转可会使指定格子以及其上下左右相邻的格子反色），并求出最优解。

解题思路是先指定第一行格子的翻转方法。并判断下一行与之相邻的格子是否需要翻转（连续翻转两次==不反转）。以此类推。判断最后一行是否全部为白色，如果不是全白则说明无解。
该算法复杂度为O(MN2^n)，符合条件。

参考代码：

#include
#include
#include
#include
using namespace std;
#define MXM 20
#define MXN 20
//邻接格子的坐标
const int dx[]={-1,0,0,0,1};
const int dy[]={0,-1,0,1,0};

int M,N;
int tile[MXN][MXN];

int opt[MXM][MXN];//保存最优解
int flip[MXM][MXN];//保存中间结果

//查询(x,y)格子的颜色
int get(int x,int y){
    int c=tile[x][y];
    for(int d=0;d>j&1;
        int num=calc();
        if(num>=0&&(resnum)){
            res=num;
            memcpy(opt,flip,sizeof(flip));
        }
    }
    if(res>M>>N)
    {
        for(int i=0; i>tile[i][j];
        solve();
    }
    return 0;
}

参考：《挑战程序设计竞赛》