POJ 3279 Fliptile

Fliptile

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 5386		Accepted: 2047

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

题目大意,：在N*M的格子里面有一些格子是黑色的，奶牛想把他全部变成白色，每次翻动一块的时候他上下左右都会一起翻面，确定再翻面最小的情况下按字典序最小输出翻的格子

思路：枚举每次枚举第一行翻面的情况，之后前一行如果出现黑色第二行就必须翻最后当最后一行全部是白色的时候就可以确定已经全部翻面为白色。

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define inf 0x3f3f3f3f
#define maxn 17
#define maxm 10000

int n,m,x,vis;
bool Map[maxn][maxn],dis[maxn][maxn],ans[maxn][maxn];
int disx[5][2] = {0,0,-1,0,0,-1,1,0,0,1};

void fun()
{
    bool M[maxn][maxn];
    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= m; j++)
            M[i][j] = Map[i][j];
    int v = 0;
    for(int j = 1; j <= m; j++)
    {
        if(dis[1][j]){
            v++;
            for(int k = 0; k < 5; k++)
                M[1+disx[k][0]][j+disx[k][1]] = !M[1+disx[k][0]][j+disx[k][1]];
        }
    }
    for(int i = 2; i <= n; i++)
        for(int j = 1; j <= m; j++)
        {
            if(M[i-1][j])
            {
                v++;
                dis[i][j] = true;
                for(int k = 0; k < 5; k++)
                    M[i+disx[k][0]][j+disx[k][1]] = !M[i+disx[k][0]][j+disx[k][1]];
            }
        }
    bool flag = true;
    for(int i = 1; i <= m; i++)
        if(M[n][i])
        {
            flag = false;
            break;
        }
    if(flag && v < vis)
    {
        vis = v;
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= m; j++ )
                ans[i][j] = dis[i][j];
    }
}

int main()
{
    vis = inf;
    scanf("%d%d",&n,&m);
    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= m; j++)
        {
            scanf("%d",&x);
            if(x) Map[i][j] = true;
            else Map[i][j] = false;
        }
    for(int i = 0; i <= (1<<m)-1 ; i++)//其实这里是可以打表节约时间的因为懒
    {
        int k = 0;
        memset(dis,0,sizeof(dis));
        for(int j = 0; j < m; j++)
        {
            if(i&(1<<j)) dis[1][j+1] = true;
        }
        fun();
    }
    if(vis == inf) printf("IMPOSSIBLE\n");
    else
    {
        for(int i = 1; i <= n; i++)
        {
            cout<<ans[i][1];
            for(int j = 2; j <= m; j++ )
                cout<<' '<<ans[i][j];
            cout<<endl;
        }
    }
    return 0;
}