版权声明:本文为博主原创文章,未经博主允许不得转载。 https://blog.csdn.net/qq_30358129/article/details/81257037
题目链接
题目大意:
给一个n*m的黑白棋盘,每次可以翻转一个块和其上下左右四个块,问最少多少次翻转出全白棋盘。
思路:枚举第一行的翻转情况,然后检验
#include <iostream>
#include <cstdio>
#include <ctime>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <cstdio>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <algorithm>
#define ll long long
#define IOS {ios::sync_with_stdio(0);cin.tie(0);}
using namespace std;
const int N = 1e5+1000;
int s[30][30];
int n,m,minn=1e9;
int out[30][30];
int rem[30][30];
void filp(int x,int y){
s[x][y] = !s[x][y];
s[x+1][y] = !s[x+1][y];
s[x-1][y] = !s[x-1][y];
s[x][y+1] = !s[x][y+1];
s[x][y-1] = !s[x][y-1];
}
void check(int x){
int f[30][30];
memset(f,0,sizeof(f));
int now = m;
int ans = 0;
while(x){
int t=x%2;
if(t){filp(1,now);f[1][now] = 1;ans++;}
x/=2;
now--;
}
int i,j;
for(i = 2;i <= n;i ++)
for(j = 1;j <= m;j ++){
if(s[i-1][j]){
filp(i,j);
f[i][j] = 1;
ans ++;
}
}
for(i = 1;i <= m;i ++)
if(s[n][i]==1) break;
if(i!=m+1)return;
if(minn>ans){
minn = ans;
memcpy(out,f,sizeof(out));
}
}
int main(){
freopen("a.txt","r",stdin);
IOS;
int i,j;
cin>>n>>m;
for(i = 1;i <= n;i ++)
for(j = 1;j <= m;j ++)
cin>>s[i][j];
memcpy(rem,s,sizeof(rem));
for(i = 0;i < (1<<m);i ++){
memcpy(s,rem,sizeof(s));
check(i);
}
if(minn==1e9){
cout<<"IMPOSSIBLE";
return 0;
}
for(i = 1;i <= n;i ++)
for(j = 1;j <= m;j ++){
cout<<out[i][j];
if(j!=m)cout<<' ';
else cout<<endl;
}
return 0;
}