POJ 2965バイナリ列挙問題

カタログ:2020年8月16日21:41:57

テーマ:

Input

Output

題意:

分析

タイトル:

The game “The Pilots Brothers: following the stripy elephant” has a quest where a player needs to open a refrigerator.
There are 16 handles on the refrigerator door. Every handle can be in one of two states: open or closed. The refrigerator is open only when all handles are open. The handles are represented as a matrix 4х4. You can change the state of a handle in any location [i, j] (1 ≤ i, j ≤ 4). However, this also changes states of all handles in row i and all handles in column j.
The task is to determine the minimum number of handle switching necessary to open the refrigerator.

Input

The input contains four lines. Each of the four lines contains four characters describing the initial state of appropriate handles. A symbol “+” means that the handle is in closed state, whereas the symbol “−” means “open”. At least one of the handles is initially closed.

Output

The first line of the input contains N – the minimum number of switching. The rest N lines describe switching sequence. Each of the lines contains a row number and a column number of the matrix separated by one or more spaces. If there are several solutions, you may give any one of them.
Sample Input

- + - -
- - - -
- - - -
- + - -

Sample Output

6
1 1
1 3
1 4
4 1
4 3
4 4

タイトル:

つの冷蔵庫に4*4の計16個のスイッチがあり、いずれかのスイッチの状態を変える(すなわち、スイッチがオフになり、スイッチがオンになる)と、このスイッチの同じ行、同じ列のすべてのスイッチが自動的に状態を変える.冷蔵庫を開けるには、すべてのスイッチを入れなければなりません.入力:1つ4×4の行列、+は閉じる、-は開く;出力:冷蔵庫を開くのに必要な最小限の操作回数と、操作したスイッチ座標.

ぶんせき

列挙による最短パスの検索

#include 
#include 
using namespace std;

typedef pair<int,int> PII;
vector<PII> path,ans;

string s;
int state = 0;
int a[4][4];

void init(){
	for(int i = 0; i < 4; i++){
		for(int j = 0; j < 4; j++){
			for(int k = 0; k < 4; k++){
				a[i][j] += 1 << (i * 4 + k);
				a[i][j] += 1 << (k * 4 + j);
			}
			a[i][j] -= 1 << (i * 4 + j);
		}
	}
}

void bp(int p){
	int t = state;
	path.clear();
	
	for(int i = 0; i < 16; i++){
		if((p >> i) & 1){
			int x = i / 4;
			int y = i % 4;
			PII tem;
			tem.first = x;
			tem.second = y;
			path.push_back(tem);
			
			t ^= a[x][y];	
		}
	}
	if(t == 0) {
		if(ans.empty() || ans.size() > path.size()){
			ans = path;
		}
	}
}

void print(){
	cout<<ans.size()<<endl;
	for(int i = 0; i < ans.size(); i++){
		cout<<ans[i].first + 1<<" "<<ans[i].second + 1<<endl;
	}
}

int main(){
	init();
	for(int i = 0; i < 4; i++){
		cin>>s;
		for(int j = 0; j < 4 ; j++){
			if(s[j] == '+') state = state+( 1<< (i*4 + j));
		} 
	}

	
	for(int p = 0; p < (1 << 16); p++) bp(p);
	
	print();
}