word cow 現在 出現 mine int pos long long sin

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

題意:有一個n*m的矩形 裏面有0/1矩陣 現在需要翻轉（每次上下左右中翻轉）成全0的矩陣 問最少需要多少次

思路：二進制枚舉第一行的所有翻轉情況 然後保證1~(m-1)行沒有1出現 這樣我們只要遍歷最後一行 只要沒有1就是一種方案

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<string>
#include<vector>
#include<stack>
#include<bitset>
#include<cstdlib>
#include<cmath>
#include<set>
#include<list>
#include<deque>
#include<map>
#include<queue>
#define ll long long int
using namespace std;
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
int moth[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
int dir[5][2]={1,0 ,0,1 ,-1,0 ,0,-1,0,0};
int dirs[8][2]={1,0 ,0,1 ,-1,0 ,0,-1, -1,-1 ,-1,1 ,1,-1 ,1,1};
const int inf=0x3f3f3f3f;
const ll mod=1e9+7;
int m,n;
int G[30][30];
int temp[30][30];
int ans[30][30];
int get(int x,int y){ //計算當前位置是否為1 
    int c=G[x][y];
    for(int i=0;i<5;i++){
        int xx=x+dir[i][0];
        int yy=y+dir[i][1];
        if(xx>=1&&xx<=m&&yy>=1&&yy<=n)
            c+=temp[xx][yy];
    }
    return c%2;
}
int solve(){
    for(int i=2;i<=m;i++)
        for(int j=1;j<=n;j++)
            if(get(i-1,j)) //如果上面一個位置是1 就需要翻轉當前位置 
                temp[i][j]=1;
    for(int i=1;i<=n;i++)
        if(get(m,i))
            return -inf;
    int cnt=0;
    for(int i=1;i<=m;i++)
        for(int j=1;j<=n;j++)
            if(temp[i][j]) cnt++;
    return cnt;
}
int main(){
    ios::sync_with_stdio(false);
    while(cin>>m>>n){
        for(int i=1;i<=m;i++)
            for(int j=1;j<=n;j++)
                cin>>G[i][j];
        int anss=-inf;
        for(int i=0;i<(1<<n);i++){ //二進制枚舉第一行的所有情況 
            memset(temp,0,sizeof(temp));
            for(int j=1;j<=n;j++)
                temp[1][j]=(i>>(j-1))&1;
            int te=solve();
            if(te>anss){
                for(int p=1;p<=m;p++)
                    for(int q=1;q<=n;q++)
                        ans[p][q]=temp[p][q];
                anss=te;
                break;
            }
        }
        if(anss==-inf) cout<<"IMPOSSIBLE"<<endl;
        else{
            for(int i=1;i<=m;i++){
                for(int j=1;j<=n;j++)
                    if(j==1) cout<<ans[i][j];
                    else cout<<" "<<ans[i][j];
                cout<<endl;
            }        
        }
    }
    return 0;
}

poj 3279 Fliptile(二進制搜索)