POJ 3279 Fliptile (暴力枚举+DFS)

本文探讨了一个复杂的黑白棋反转问题,目标是最小化翻转次数使所有棋子变为白色面朝上,考虑到每次翻转会同时影响周围棋子,提供了一种解决策略并附带源代码实现。

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p.s. 黑白棋反转问题

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

p.s. 题目大意:给你一个M*N的棋盘,1代表翻转,0代表不翻转,问你怎样才能反转最少的次数让1都变成0,你的每次翻转都会翻转它的上下左右棋子,最后输出你翻转的方式,如果有多种方式,输出字典序最小的,如果不能完成翻转,输出"IMPOSSIBLE"

#include <iostream>
#include <string.h>

using namespace std;

struct node{
    int q,a,z,t;
}now,tmp;

int m,n;
int Map[17][17];
int ans[17][17];
int f[17][17];
int Min;

void dfs(int x,int num)//x代表第x行,num代表翻转次数
{
    if(num>Min)//超过最小翻转次数,舍去此方案
    {
        return;
    }
    if(x>n)//跑完第n行
    {
        int flag=1;
        for(int i=1;i<=m;i++)//判断第n行是否都被翻转
        {
            if((Map[n][i]+f[n][i]+f[n][i-1]+f[n][i+1]+f[n-1][i])%2)
            {
                flag=0;
            }
        }
        if(flag&&num<Min)//如果第n行都被翻转,并且此时num小于最小翻转次数,更新翻转方案
        {
            memcpy(ans,f,sizeof(f));
            Min=num;
        }
        return;
    }
    int t=0;
    for(int i=1;i<=m;i++)
    {
        if((Map[x-1][i]+f[x-2][i]+f[x-1][i-1]+f[x-1][i+1]+f[x-1][i])%2)//判断[x-1][i]是否为1
        {
            f[x][i]=1;
            t++;
        }
        else
        {
            f[x][i]=0;
        }
    }
    dfs(x+1,num+t);
}

void Enum(int y,int num)//y代表第1行第y个棋子,num代表翻转次数,枚举第一行所有情况
{
    if(y>m)
    {
        dfs(2,num);//调用dfs判断第二行以后需要翻转次数
        return;
    }
    f[1][y]=0;
    Enum(y+1,num);
    f[1][y]=1;
    Enum(y+1,num+1);
}

int main()
{
    while(cin >> n >> m)
    {
        Min=0x7f7f7f7f;
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=m;j++)
            {
                cin >> Map[i][j];
            }
        }
        Enum(1,0);
        if(Min!=0x7f7f7f7f)
        {
            for(int i=1;i<=n;i++)
            {
                for(int j=1;j<=m;j++)
                {
                    cout << ans[i][j];
                    if(j!=m)
                    {
                        cout << ' ';
                    }
                    else
                    {
                        cout << endl;
                    }
                }
            }
        }
        else
        {
            cout << "IMPOSSIBLE" << endl;
        }
    }
    return 0;
}

 

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