POJ3279 （二進制枚舉）

題意描述

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”.

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

思路

我們只需要枚舉出第一行的狀態，然後根據第一行的狀態來遞推出其餘每一行的狀態，在遞推完以後，還需要判斷最後一行燈是否滿足題意。我們可以不在原數組中改變狀態，而是根據該位置的狀態是否改變過，從而絕對是否對該位置進行操作，這樣可以大大的節省時間。方法1是在原數組中改變，方法2是根據位置的狀態來決定是否操作。

AC代碼

方法1：使用G++編譯器可以勉強通過

#include<iostream>
#include<algorithm>
#include<vector>
#include<ctime>
#include<cstring>
#define IOS ios::sync_with_stdio(false); cin.tie(0); 
using namespace std;
typedef pair<int,int> PII;
const int N=20;
const int INF=0x3f3f3f3f;
int n,m;
int g[N][N],backup[N][N];
int ans[N][N];
int dx[5]={1,0,-1,0,0},dy[5]={0,1,0,-1,0};
void turn(int x,int y){
	for(int i=0;i<5;i++){
		int gx=x+dx[i],gy=y+dy[i];
		if(gx>=0&&gy>=0&&gx<n&&gy<m){
			g[gx][gy]^=1;
		}
	}
}
int main(){
	IOS;
	cin>>n>>m;
	for(int i=0;i<n;i++){
		for(int j=0;j<m;j++){
			cin>>g[i][j];
		}
	}
	//clock_t starttime,endtime;	
	//starttime=clock();
	int ccc=INF;
	for(int op=0;op<1<<m;op++){
		int cnt=0;
		memcpy(backup,g,sizeof g);
		vector<PII> res;
		PII buf;
		for(int i=0;i<m;i++){
			if(op>>i&1){
				turn(0,i);
				cnt++;
				buf.first=0,buf.second=i;
				res.push_back(buf);
			}
		}
		for(int i=1;i<n;i++){
			for(int j=0;j<m;j++){
				if(g[i-1][j]==1){
					turn(i,j);
					buf.first=i,buf.second=j;
					res.push_back(buf);
					cnt++;
				}
			}
		}
		bool is_closed=true;
		for(int i=0;i<m;i++){
			if(g[n-1][i]==1){
				is_closed=false;
				break;
			}
		}
		if(is_closed){
			if(cnt<ccc){
				ccc=cnt;
				memset(ans,0,sizeof ans);
				for(int i=0;i<res.size();i++){
					int gx=res[i].first,gy=res[i].second;
					ans[gx][gy]++;
				}
			}
		}
		memcpy(g,backup,sizeof backup);
	}
	if(ccc==INF){
		cout<<"IMPOSSIBLE"<<endl;
	}else{
		for(int i=0;i<n;i++){
			for(int j=0;j<m;j++){
				cout<<ans[i][j]<<' ';
			}
			cout<<endl;
		}
	}
	//endtime=clock();
	//cout<<(double)(endtime - starttime) / CLOCKS_PER_SEC<<"s"<<endl;
	return 0;
}

方法2：

#include<iostream>
#include<algorithm>
#include<vector>
#include<ctime>
#include<cstring>
#define IOS ios::sync_with_stdio(false); cin.tie(0); 
using namespace std;
typedef pair<int,int> PII;
const int N=20;
const int INF=0x3f3f3f3f;
int n,m;
int g[N][N];
int ans[N][N];
int choose[N][N];
int dx[5]={1,0,-1,0,0},dy[5]={0,1,0,-1,0};
int get(int x,int y){
	int temp=g[x][y];
	for(int i=0;i<5;i++){
		int gx=x+dx[i],gy=y+dy[i];
		if(gx>=0&&gy>=0&&gx<n&&gy<m){
			temp+=choose[gx][gy];
		}
	}
	return temp&1;
}
int main(){
	IOS;
	cin>>n>>m;
	for(int i=0;i<n;i++){
		for(int j=0;j<m;j++){
			cin>>g[i][j];
		}
	}
	//clock_t starttime,endtime;	
	//starttime=clock();
	int ccc=INF;
	for(int op=0;op<1<<m;op++){
		int cnt=0;
		memset(choose,0,sizeof choose);
		for(int i=0;i<m;i++){
			choose[0][m-i-1]=op>>i&1;
		}
		for(int i=1;i<n;i++){
			for(int j=0;j<m;j++){
				if(get(i-1,j)){
					choose[i][j]=1;
				}
			}
		}
		bool is_closed=true;
		for(int i=0;i<m;i++){
			if(get(n-1,i)){
				 is_closed=false;
				 break;
			}
		}
		for(int i=0;i<n;i++){
			for(int j=0;j<m;j++){
				cnt+=choose[i][j];
			}
		}
		if(is_closed){
			if(cnt<ccc){
				ccc=cnt;
				memcpy(ans,choose,sizeof choose);
			}
		}
	}
	if(ccc==INF){
		cout<<"IMPOSSIBLE"<<endl;
	}else{
		for(int i=0;i<n;i++){
			for(int j=0;j<m;j++){
				cout<<ans[i][j]<<' ';
			}
			cout<<endl;
		}
	}
	return 0;
}