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  | #include<stdio.h>
#include<algorithm>
#include<math.h>
#include<string.h>
using namespace std;
const int MAXN = 1005;
int L[MAXN*MAXN], R[MAXN*MAXN], U[MAXN*MAXN], D[MAXN*MAXN];
int S[MAXN];
int Col[MAXN*MAXN], Row[MAXN*MAXN],ans,limit,up;
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]) {
		U[D[j]] = j;
		D[U[j]] = j;
		++ S[Col[j]];
	}
	L[R[c]] = c;
	R[L[c]] = c;
}
bool dfs(int depth) {
	if (depth >= ans) return true;
 
	if(R[0] == 0) { if(depth < ans)ans = depth;  return true; }
 
	int i, j, c, minnum = 2000000000, flag = 0;
	for (i = R[0]; i != 0; i = R[i]) {
		if (S[i] < minnum) {
			minnum = S[i];
			c = i;
		}
	}
	Remove(c);
	for (i = U[c]; i != c; i = U[i]) {
		//如果需要的话,在这里记录一组解(Ans[depth] = Row[i])
		for (j = R[i]; j != i; j = R[j]) Remove(Col[j]);
		if (dfs(depth + 1)) flag = 1; //return true;
		for (j = L[i]; j != i; j = L[j]) Resume(Col[j]);
	}
	Resume(c);
	return flag;//false;
}
int solve(int n, int m, int DL[][MAXN]) {
	for (int i = 0; i <= m; i ++) {
		L[i] = i - 1;
		R[i] = i + 1;
		U[i] = D[i] = i;
	}
	L[0] = m;
	R[m] = 0;
	int cnt = m + 1;
	memset(S, 0, sizeof (S));
	for (int i = 1; i <= n; i ++) {
		int head = cnt, tail = cnt;
		for (int c = 1; c <= m; c ++) if (DL[i][c]) {
			S[c] ++;
			Row[cnt] = i;
			Col[cnt] = c;
			U[D[c]] = cnt;
			D[cnt] = D[c];
			U[cnt] = c;
			D[c] = cnt;
			L[cnt] = tail;
			R[tail] = cnt;
			R[cnt] = head;
			L[head] = cnt;
			tail = cnt;
			cnt ++;
		}
	}
	if (dfs(0)) return true;
	return false;
}
int mark[MAXN][MAXN];
int main()
{
	int i,j,k,l,n,m,T,row,col,x1,x2,y1,y2,id[33][33],low;
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d %d %d",&n,&m,&row);
		col = 0;
		for(i=1;i<=n;i++)
			for(j=1;j<=m;j++)
			{
				col++; id[i][j] = col;
			}
 
		for(i=1;i<=row;i++)
			for(j=1;j<=col;j++)mark[i][j] = 0;
		for(i=1;i<=row;i++)
		{
			scanf("%d %d %d %d",&x1,&y1,&x2,&y2);
			x1++; y1++;
			for(k=x1;k<=x2;k++)
				for(l=y1;l<=y2;l++)
				mark[i][id[k][l]] = 1;
		}
		ans = 1000000000;
		if(!solve(row, col, mark))ans = -1;
		printf("%d\n",ans);
	}
	return 0;
}
  |