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 n m p (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 x1 y1 x2 y2 (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.
题意：告诉你p个地图碎片，让你拼成一个n*m的地图，碎片之间不能重合。碎片不必用完。
DLX做法：n*m的大小分成n*m个方格作为M，p个地图碎片作为N，每个地图碎片可以覆盖若干个方格.
于是就转化成了DLX的精确覆盖问题。套一个DLX精确覆盖模板即可。
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<string>
#include<iostream>
#include<queue>
#include<cmath>
#include<map>
#include<stack>
#include<bitset>
using namespace std;
#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define CLEAR( a , x ) memset ( a , x , sizeof a )
typedef long long LL;
typedef pair<int,int>pil;
const int maxn = 1100;
const int maxnode = 500010;
const int mod = 1000000007;
struct DLX
{
int n,m,size;
int U[maxnode],D[maxnode],R[maxnode],L[maxnode],Row[maxnode],Col[maxnode];
int H[maxn],S[maxn];
int ansd;
void init(int a,int b)
{
n = a;
m = b;
for(int i = 0;i <= m;i++)
{
S[i] = 0;
U[i] = D[i] = i;
L[i] = i-1;
R[i] = i+1;
}
R[m] = 0; L[0] = m;
size = m;
for(int i = 1;i <= n;i++)
H[i] = -1;
}
void Link(int r,int c)
{
++S[Col[++size]=c];
Row[size] = r;
D[size] = D[c];
U[D[c]] = size;
U[size] = c;
D[c] = size;
if(H[r] < 0)H[r] = L[size] = R[size] = size;
else
{
R[size] = R[H[r]];
L[R[H[r]]] = size;
L[size] = H[r];
R[H[r]] = size;
}
}
void remove(int c)
{
L[R[c]] = L[c]; R[L[c]] = R[c];
for(int i = D[c];i != c;i = D[i])
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 c)
{
for(int i = U[c];i != c;i = U[i])
for(int j = L[i];j != i;j = L[j])
++S[Col[U[D[j]]=D[U[j]]=j]];
L[R[c]] = R[L[c]] = c;
}
void Dance(int d)
{
if(ansd!=-1&&d>ansd) return ;
if(R[0]==0)
{
if(ansd==-1) ansd=d;
else if(ansd>d) ansd=d;
return ;
}
int c=R[0];
for(int i=R[0];i!=0;i=R[i])
{
if(S[i]<S[c])//选择1的数量最少的开始搜
c=i;
}
remove(c);
for(int i=D[c];i!=c;i=D[i])
{
for(int j=R[i];j!=i;j=R[j]) remove(Col[j]);
Dance(d+1);
for(int j=L[i];j!=i;j=L[j]) resume(Col[j]);
}
resume(c);
}
};
int t,n,m,p;
DLX L;
int main()
{
int x1,y1,x2,y2;
scanf("%d",&t);
while(t--)
{
scanf("%d%d%d",&n,&m,&p);
L.init(p,n*m);
REPF(k,1,p)
{
scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
for(int i=x1+1;i<=x2;i++)
{
for(int j=y1+1;j<=y2;j++)
L.Link(k,j+(i-1)*m);
}
}
L.ansd=-1;
L.Dance(0);
printf("%d\n",L.ansd);
}
return 0;
}