传送门POJ3279

描述

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”.

输入

Line 1: Two space-separated integers: M and NLines 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

输出

Lines 1..M: Each line contains N space-separated integers, each specifying how many times to flip that particular location.

样例

Input4 41 0 0 10 1 1 00 1 1 01 0 0 1

Output0 0 0 01 0 0 11 0 0 10 0 0 0

题解

题意：翻牌子，每翻一次可以将选中的牌子自身以及上下左右取反，求最少多少次可以翻成全0.对于第i行，我们可以找到i-1行中为1的位置[i-1][j]，然后通过翻[i][j]将其变为0，这样可以保证除了最后一行，前面全是0，最后只需要看最后一行是否为全0就可以判断是否可行。枚举第一行，共2^N种，通过这些初始状态去处理后面的格子，如果可行，更新答案。

Code

1234567891011121314151617181920212223242526272829303132333435363738394041424344454647484950515253545556575859606162636465666768697071#include<iostream>#include<cstring>#define INIT(a,b) memset(a,b,sizeof(a))#define LL long longusing namespace std;const int inf=0x3f3f3f3f;const int N=1e2+7;const int mod=1e9+7;int m,n,mp[N][N],tem[N][N],ans[N][N],res[N][N],flag=0,time,mintime=inf;void init(){ for(int i=0;i<m;i++) for(int j=0;j<n;j++) tem[i][j]=mp[i][j]; INIT(ans,0); time=0;}void fan(int i,int j){ tem[i][j]^=1; if(i-1>=0) tem[i-1][j]^=1; if(i+1<m) tem[i+1][j]^=1; if(j-1>=0) tem[i][j-1]^=1; if(j+1<n) tem[i][j+1]^=1;}bool ac(){ int i; for(i=0;i<n;i++) if(tem[m-1][i]) break; if(i==n) return true; return false;}int main(){ ios::sync_with_stdio(false); cin.tie(0); cin>>m>>n; for(int i=0;i<m;i++) for(int j=0;j<n;j++) cin>>mp[i][j]; for(int sta=0;sta<(1<<n);sta++){ //状态 init(); for(int j=0;j<n;j++){ //处理第一行 if(sta & (1<<j)){ ans[0][n-j-1]=1; fan(0,n-j-1); time++; } } for(int i=1;i<m;i++){ for(int j=0;j<n;j++){ if(tem[i-1][j]){ ans[i][j]=1; fan(i,j); time++; } } } if(ac() && time<mintime){ mintime=time; for(int i=0;i<m;i++) for(int j=0;j<n;j++) res[i][j]=ans[i][j]; } } if(mintime==inf) cout<<"IMPOSSIBLE"<<endl; else { for(int i=0;i<m;i++) for(int j=0;j<n;j++) cout<<res[i][j]<<" \n"[j==n-1]; } return 0;}