poj3279 Fliptile 开关问题

Fliptile

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 12725 Accepted: 4681

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 41 0 0 10 1 1 00 1 1 01 0 0 1

Sample Output

0 0 0 01 0 0 11 0 0 10 0 0 0

题意：翻棋盘，每次翻棋子将带动上下左右棋子一起翻。求将所有棋子都翻成白色的最小着法。

分析：如果将整个棋盘用一个数来表示的话，需要2的m*n次方位数，225位，显然不行。可以枚举第一行的排序，有2的n次方种方式，然后依次往下一行推进。m-1行，每行需要遍历n次，因此复杂度为2^n*m*n。

#include <cstdio>#include <cstdlib>#include <iostream>#include <stack>#include <queue>#include <algorithm>#include <cstring>#include <string>#include <cmath>#include <vector>#include <bitset>#include <list>#include <sstream>#include <set>#include <functional>using namespace std;#define INF 0x3f3f3f3f#define maxn 15typedef long long ll;int M,N;int inp[maxn][maxn];int mid[maxn][maxn];int best[maxn][maxn];int dx[5] = {-1,1,0,0,0};int dy[5] = {0,0,0,1,-1};int get(int x,int y){int c = inp[x][y];for (int i = 0; i < 5; ++i){int nx = x+dx[i],ny = y+dy[i];if(nx>=0 && nx<M && ny>=0 && ny<N) c += mid[nx][ny];}return c%2;//返回1说明是黑色，需要翻转}int cal(){for (int i = 1; i < M; ++i)for (int j = 0; j < N; ++j)mid[i][j] = get(i-1,j);for (int i = 0; i < N; ++i)if(get(M-1,i)) return -1;int res = 0;for (int i = 0; i < M; ++i)for (int j = 0; j < N; ++j)res += mid[i][j];return res;}void solve(){int ans = -1;for (int i = 0; i < 1<<N; ++i){//第一行每次翻转memset(mid,0,sizeof(mid));for (int j = 0; j < N; ++j)mid[0][N-1-j] = ((i>>j) & 1);int sum = cal();//计算此次翻转的次数if(sum>=0 && (ans==-1 || ans>sum)) ans = sum,memcpy(best,mid,sizeof(mid));}if(ans == -1) printf("IMPOSSIBLE\n");else{for (int i = 0; i < M; ++i){for (int j = 0; j < N; ++j)cout << best[i][j] << " ";cout << endl;}}}int main(int argc, char const *argv[]){#ifndef ONLINE_JUDGEfreopen("in.txt","r",stdin);freopen("out.txt","w",stdout);#endifcin >> M >> N;for (int i = 0; i < M; ++i)for (int j = 0; j < N; ++j)cin >> inp[i][j];solve();return 0;}