POJ 3279 Fliptile

传送门：http://poj.org/problem?id=3279

Fliptile

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 8322		Accepted: 3102

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

//😭😭现在还不会做，慢慢看吧，在网上看了一下他们的解题报告
1.如果处理了前面n-1行了，我们只要判断最后一行是否是全为0
2.现在才明白怎们用暴力方法，每一个格子都有两种操作 翻还是不翻，这样就有2^(M+N)种可能性，然后一次枚举每一种，找出翻转最少的就是它的解。
3.还是不知道怎么弄，以后持续更新吧。
现在完全明白了，和POJ 3276 Face The Right Way  http://poj.org/problem?id=3276 类似，上面这个简单点，
这类问题统称为开关问题，它注重开头和结尾，开头决定了，后面的状态也可由前面的状态推出来，而判断末尾情况，可以发现该翻转方式是否可行。

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

const int MAX_N=100;
const int dx[]={-1,1,0,0,0};
const int dy[]={0,0,-1,0,1};

int flip[MAX_N][MAX_N],tile[MAX_N][MAX_N],opt[MAX_N][MAX_N];
int N,M;

//这是判断该瓷砖是否需要翻转

int get(int x,int y){
    int c=tile[x][y];

    for(int i=0;i<5;i++){
        int tx=x+dx[i],ty=y+dy[i];
        if(tx>=0&&tx<M&&ty>=0&&ty<N)
            c+=flip[tx][ty];
    }
    return c%2;
}

int calc(){
    for(int i=1;i<M;i++)
    for(int j=0;j<N;j++)
    if(get(i-1,j)&1){    //如何判断[i][j]瓷砖是否需要翻转呢，这是判断它的上一个[i-1][j]是否是1
        flip[i][j]=1;
    }

    //检测最后一行是否全为0，如果不是则该翻转不行。
    for(int j=0;j<N;j++){
        if(get(M-1,j)&1)
            return -1;
    }
    //这是计算翻转了多少次
    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;
    //二进制枚举第一行的所有选项
    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)){
            memcpy(opt,flip,sizeof(flip));
            res=num;
        }
    }

    if(res<0) 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(){
    cin>>M>>N;
    for(int i=0;i<M;i++)
        for(int j=0;j<N;j++)
        scanf("%d",&tile[i][j]);
    solve();
}