POJ - 3279 Fliptile （状压+暴力）

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

Sample Output

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

题目大意：n*m的图，要把所有的1都翻成0至少要翻几次，其中每次翻一个数它的上下左右都会被翻

思路：枚举第一行的所有翻法，只要确定第一行的翻法那下面就都确定了，因为下面只需要保证把上面的那行的1全翻成0

枚举的时候每个位置只有翻和不翻两种状态，所以可以用状压的思想解决

代码：

#include
#include
#include
using namespace std;
#define inf 0x3f3f3f3f
int mp[20][20];
int c[20][20],a[20][20],ma[20][20];
int n,m,ans,go[5][2]= {0,1,1,0,-1,0,0,-1,0,0};
void filp(int x,int y)
{
    a[x][y]=1;
    for(int i=0; i<5; i++)
    {
        int xx=x+go[i][0],yy=y+go[i][1];
        c[xx][yy]=!c[xx][yy];
    }
}
int isEmp()
{
    for(int i=1; i<=m; i++)
        if(c[n][i]==1) return 0;
    return 1;
}
void solve(int x)
{
    memcpy(c,mp,sizeof(mp));
    memset(a,0,sizeof(a));
    int cnt=0;
    for(int i=0; i>i&1;
        if(x>>i&1)
        {
            filp(1,i+1);
            cnt++;
        }
    }
    for(int i=2; i<=n; i++)
        for(int j=1; j<=m; j++)
        {
            if(c[i-1][j]==1)
            {
                filp(i,j);
                cnt++;
            }
        }
    if(isEmp()&&cnt=inf) printf("IMPOSSIBLE\n");
        else
        {
            for(int i=1; i<=n; i++)
                for(int j=1; j<=m; j++)
                    printf("%d%c",ma[i][j],j==m?'\n':' ');
        }
    }
    return 0;
}