演算法學習-Dancing Links X

Dancing Links X 舞蹈鏈。
實質為用迴圈十字鏈在圖上將所有“1”的位置鏈起來
構造較為巧妙，且極易理解，本題為 DLX 模板（精確覆蓋問題）
DLX 演算法的題目做法一般為將所求方案轉化為行號，將限制條件轉化為列號
然後 dfs 暴力列舉，用迴圈十字鏈最佳化

/*
Finished at 12:14 on 2024.4.5
*/
#include <bits/stdc++.h>

using namespace std;

const int N = 510, M = 250510;

int n, m;
int a[N][N];
int row[M], col[M];
//row為每個點所在行號，col為每個點所在列號 
int cnt, s[N], h[N];
//cnt為給每個鏈上的點的編號
//s表示某一列上所建連結串列點個數
//h為每一行的列表頭 
int u[M], d[M], l[M], r[M];
//u, d, l, r分別表示某個連結串列點上下左右所連的 
int res[N];
//所選行號 

void init()              //初始化第0行的連結串列頭 
{
	for (int i = 0; i <= m; i ++ )
		u[i] = d[i] = i, l[i] = i - 1, r[i] = i + 1;     //初始化左右，上下還沒點，所以指向自己 
	l[0] = m, r[m] = 0, cnt = m;                        //處理剩下的0，m點 
}

void link(int x, int y)
{
	s[y] ++ ;
	cnt ++ ;
	row[cnt] = x, col[cnt] = y;
	u[cnt] = y;
	d[cnt] = d[y];               //可類比連結串列，正常加即可 
	u[d[y]] = cnt;
	d[y] = cnt;
	if (!h[x]) h[x] = l[cnt] = r[cnt] = cnt;     //本行無連結串列點，則加進去 
	else
	{
		l[cnt] = l[h[x]];
		r[cnt] = h[x];                           //正常雙向連結串列加 
		r[l[h[x]]] = cnt;
		l[h[x]] = cnt;
	}
}

void remove(int x)
{
	r[l[x]] = r[x], l[r[x]] = l[x];
	for (int i = d[x]; i != x; i = d[i])                  //向下，向右刪除每個點 
		for (int j = r[i]; j != i; j = r[j])
			u[d[j]] = u[j], d[u[j]] = d[j], s[col[j]] -- ;
}

void resume(int x)
{
	r[l[x]] = x, l[r[x]] = x;
	for (int i = u[x]; i != x; i = u[i])                  //向上，向左恢復每個點 
		for (int j = l[i]; j != i; j = l[j])
			u[d[j]] = j, d[u[j]] = j, s[col[j]] ++ ;
}

bool dance(int depth)
{
	if (r[0] == 0)
	{
		for (int i = 0; i < depth; i ++ ) cout << res[i] << ' ';
		cout << '\n';              //第0行刪完了 
		return true;
	}
	
	int y = r[0];
	for (int i = r[0]; i; i = r[i])    //優先找1少的 
		if (s[y] > s[i]) y = i;
	
	remove(y);
	for (int i = d[y]; i != y; i = d[i])
	{
		res[depth] = row[i];
		for (int j = r[i]; j != i; j = r[j]) remove(col[j]);
		if (dance(depth + 1)) return true;                          //暴力列舉 
		for (int j = l[i]; j != i; j = l[j]) resume(col[j]);
	}
	resume(y);
	
	return false;
}

int main()
{
	cin >> n >> m;
	for (int i = 1; i <= n; i ++ )
		for (int j = 1; j <= m; j ++ )
			cin >> a[i][j];
	
	init();
	for (int i = 1; i <= n; i ++ )
		for (int j = 1; j <= m; j ++ )
			if (a[i][j]) link(i, j);          //1位置加點 
	
	if (!dance(0)) cout << "No Solution!\n";
			
	return 0;
}