POJ - 3279 Fliptile

POJ - 3279 Fliptile

题目

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

题解

看最后一行是否满足题意即可

/*************************************************************************
    > Problem: POJ - 3279 Fliptile
    > Author: ALizen_
    > Mail: hhu_zyh@qq.com
    > Blog: https://blog.youkuaiyun.com/Jame_19?spm=1001.2101.3001.5343
*************************************************************************/
#include<set>
#include<map>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define re register int
#define sf(x) scanf("%d",&x)
#define pf(x) printf("%d\n",x)
#define slf(x) scanf("%lld",&x)
#define plf(x) printf("%lld\n",x)
#define MEM(a,b) memset((a),(b),sizeof(a))
#define ref(i,a,b) for(re i = a ; i >= b ; -- i)
#define fer(i,a,b) for(re i = a ; i <= b ; ++ i)
using namespace std;
typedef long long LL;
#define INF 0x3f3f3f3f
bool a[20][20];   //存图
bool b[20][20];   //保存当前翻转方案
bool c[20][20];   //保存最优解
int d[5][2] = { { -1, 0 }, { 1, 0 }, { 0, 0 }, { 0, -1 }, { 0, 1 } };
int ans = INF;
int n, m;
bool getcolor(int x, int y)    //得到x,y的颜色
{
	int res = a[x][y];
	for (int i = 0; i < 5; i++)
	{
		int fx = x + d[i][0], fy = y + d[i][1];
		if (fx >= 1 && fx <= n && fy >= 1 && fy <= m) res += b[fx][fy];
	}
	return res % 2;
}
int solve()
{
	int res = 0;
	for (int i = 2; i <= n; i++)    //从第二行开始检查是否需要翻转
		for (int j = 1; j <= m; j++)
			if (getcolor(i - 1, j)) b[i][j] = 1;
	for (int i = 1; i <= m; i++)     //检查最后一行是否全为白色
		if (getcolor(n, i)) return INF;
	for (int i = 1; i <= n; i++)    //统计翻转次数 
		for (int j = 1; j <= m; j++)
			res += b[i][j];
	return res;
}
int main()
{
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= m; j++)
			scanf("%d", &a[i][j]);
	for (int s = 0; s < 1 << m; s++)   //按照字典序枚举第一行所以翻转可能
	{
		memset(b, false, sizeof b);
		for (int i = 1; i <= m; i++)
			b[1][i] = s >> (m - i) & 1;
		int t = solve();
		if (t < ans)
		{
			ans = t;
			memcpy(c, b, sizeof b);
		}
	}
	if (ans == INF) printf("IMPOSSIBLE\n");
	else 
	{
		for (int i = 1; i <= n; i++)
		{
			for (int j = 1; j <= m; j++)
				printf("%d ", c[i][j]);
			printf("\n");
		}
	}
	return 0;
}