ZOJ 3209 Treasure Map

Description

Your boss once had got many copies of a treasure map. Unfortunately, all the copies are now broken to many rectangular pieces, and what make it worse, he has lost some of the pieces. Luckily, it is possible to figure out the position of each piece in the original map. Now the boss asks you, the talent programmer, to make a complete treasure map with these pieces. You need to make only one complete map and it is not necessary to use all the pieces. But remember, pieces are not allowed to overlap with each other (See sample 2).

Input

The first line of the input contains an integer T (T <= 500), indicating the number of cases.

For each case, the first line contains three integers nmp (1 <= n, m <= 30, 1 <= p <= 500), the width and the height of the map, and the number of pieces. Then p lines follow, each consists of four integers x1y1x2y2 (0 <= x1< x2 <= n, 0 <= y1 < y2 <= m), where (x1, y1) is the coordinate of the lower-left corner of the rectangular piece, and (x2, y2) is the coordinate of the upper-right corner in the original map.

Cases are separated by one blank line.

Output

If you can make a complete map with these pieces, output the least number of pieces you need to achieve this. If it is impossible to make one complete map, just output -1.

Sample Input

3
5 5 1
0 0 5 5

5 5 2
0 0 3 5
2 0 5 5

30 30 5
0 0 30 10
0 10 30 20
0 20 30 30
0 0 15 30
15 0 30 30

Sample Output

1
-1
2

Hint

For sample 1, the only piece is a complete map.

For sample 2, the two pieces may overlap with each other, so you can not make a complete treasure map.

For sample 3, you can make a map by either use the first 3 pieces or the last 2 pieces, and the latter approach one needs less pieces.

dlx精确覆盖

#include<cstdio>
#include<vector>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const ll maxn = 25;
int T, n, m, x, y, t, X, Y;

inline void read(int &ret)
{
	char c;
	do {
		c = getchar();
	} while (c < '0' || c > '9');
	ret = c - '0';
	while ((c = getchar()) >= '0' && c <= '9')
		ret = ret * 10 + (c - '0');
}

struct DLX
{
	#define maxn 500005
	#define F(i,A,s) for (int i=A[s];i!=s;i=A[i])
	int L[maxn], R[maxn], U[maxn], D[maxn];
	int row[maxn], col[maxn], ans[maxn], cnt[maxn];
	int n, m, num, sz;

	void add(int now, int l, int r, int u, int d, int x, int y)
	{
		L[now] = l;	R[now] = r;	U[now] = u;
		D[now] = d;   row[now] = x;  col[now] = y;
	}
	void reset(int n, int m)
	{
		num = 0x7FFFFFFF;
		this->n = n;	this->m = m;
		for (int i = 0; i <= m; i++)
		{
			add(i, i - 1, i + 1, i, i, 0, i);
			cnt[i] = 0;
		}
		L[0] = m; 	R[m] = 0; 	sz = m + 1;
	}
	void insert(int x, int y)
	{
		int ft = sz - 1;
		if (row[ft] != x)
		{
			add(sz, sz, sz, U[y], y, x, y);
			U[D[sz]] = sz; D[U[sz]] = sz;
		}
		else
		{
			add(sz, ft, R[ft], U[y], y, x, y);
			R[L[sz]] = sz; L[R[sz]] = sz;
			U[D[sz]] = sz; D[U[sz]] = sz;
		}
		++cnt[y];	++sz;
	}

	//精确覆盖
	void remove(int now)
	{
		R[L[now]] = R[now];
		L[R[now]] = L[now];
		F(i, D, now) F(j, R, i)
		{
			D[U[j]] = D[j];
			U[D[j]] = U[j];
			--cnt[col[j]];
		}
	}
	void resume(int now)
	{
		F(i, U, now)	F(j, L, i)
		{
			D[U[j]] = j;
			U[D[j]] = j;
			++cnt[col[j]];
		}
		R[L[now]] = now;
		L[R[now]] = now;
	}
	void dfs(int x)
	{
		if (x + A() >= num) return;
		if (!R[0]) { num = min(num, x); return ; }
		int now = R[0];
		F(i, R, 0) if (cnt[now]>cnt[i]) now = i;
		remove(now);
		F(i, D, now)
		{
			//ans[x] = row[i];
			F(j, R, i) remove(col[j]);
			dfs(x + 1);
			F(j, L, i) resume(col[j]);
		}
		resume(now);
	}
	//精确覆盖

	//重复覆盖
	void Remove(int now)
	{
		F(i, D, now)
		{
			L[R[i]] = L[i];
			R[L[i]] = R[i];
		}
	}
	void Resume(int now)
	{
		F(i, U, now) L[R[i]] = R[L[i]] = i;
	}
	int vis[maxn];
	int flag[maxn];
	int A()
	{
		int dis = 0;
		F(i, R, 0) vis[i] = 0;
		F(i, R, 0) if (!vis[i])
		{
			dis++;	vis[i] = 1;
			F(j, D, i) F(k, R, j) vis[col[k]] = 1;
		}
		return dis;
	}
	void Dfs(int x)
	{
		if (!R[0]) num = min(num, x);
		else if (x + A()<num)
		{
			int now = R[0];
			F(i, R, 0) if (cnt[now]>cnt[i]) now = i;
			F(i, D, now)
			{
				Remove(i); F(j, R, i) Remove(j);
				Dfs(x + 1);
				F(j, L, i) Resume(j); Resume(i);
			}
		}
	}
	//重复覆盖
}dlx;

int main()
{
	read(T);
	while (T--)
	{
		scanf("%d%d%d", &n, &m, &t);
		dlx.reset(t, n * m);
		for (int k = 1; k <= t; k++)
		{
			scanf("%d%d%d%d", &x, &y, &X, &Y);
			for (int i = x; i < X; i++)
				for (int j = y; j < Y; j++)
					dlx.insert(k, i * m + j + 1);
		}
		dlx.dfs(0);
		if (dlx.num <= t) printf("%d\n", dlx.num);
		else printf("-1\n");
	}
	return 0;
}