POJ3279 Fliptile

POJ3279 Fliptile

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

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2
3
4
5
4 4
1 0 0 1
0 1 1 0
0 1 1 0
1 0 0 1

Sample Output

1
2
3
4
0 0 0 0
1 0 0 1
1 0 0 1
0 0 0 0

Answer

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#include <cstdio>
#include <cstring>

int M, N;
const int dx[5] = {-1, 0, 0, 0, 1};
const int dy[5] = {0, -1, 0, 1, 0};
int ma[20][20], f[20][20], opt[20][20];

int get(int x, int y);
int calc();

int main() {
scanf("%d %d", &M, &N);
for (int i = 0; i < M; ++i)
for (int j = 0; j < N; ++j)
scanf("%d", &ma[i][j]);

int res = -1;
for (int i = 0; i < 1 << N; ++i) {
memset(f, 0, sizeof(f));
for (int j = 0; j < N; ++j) {
f[0][N-j-1] = i >> j & 1;
}
int num = calc();
if (num >= 0 && (res < 0 || res > num)) {
res = num;
memcpy(opt, f, sizeof(f));
}
}

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' : ' ');
}
return 0;
}

int calc() {
for (int i = 1; i < M; ++i) {
for (int j = 0; j < N; ++j) {
if (get(i-1, j) != 0) {
f[i][j] = 1;
}
}
}

for (int j = 0; j < N; ++j) {
if (get(M-1, j) != 0) {
return -1;
}
}

int res = 0;
for (int i = 0; i < M; ++i) {
for (int j = 0; j < N; ++j) {
res += f[i][j];
}
}
return res;
}

int get(int x, int y) {
int c = ma[x][y];
for (int d = 0; d < 5; ++d) {
int nx = x+dx[d], ny = y+dy[d];
if (0 <= nx && nx < M && 0 <= ny && ny < N) {
c += f[nx][ny];
}
}
return c % 2;
}

Analysis

  1. 每次翻转使得其上下左右相邻格子也翻转,求最少的次数使得所有各自都变为 0,次数相同,求翻转方法字典序最小的那个,不存在输出 impossible。
  2. 固定第一行的翻转方法,那么第一行就只受第二行对应位置的影响,通过枚举第一行的翻转方法,同时遍历整个数组,求出此次需要翻转的次数。
  3. 若最后一行在操作完全为 0,则成功,否则失败。
  4. 翻转受影响次数为偶数时,相当于不翻转。
# POJ

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