用最少的矩阵覆盖n*m的地图。注意矩阵不能互相覆盖。

这里显然是一个精确覆盖，但因为矩阵拼接过程中，有公共的边，这里须要的技巧就是把矩阵的左边和以下截去一个单位。

#include <stdio.h>

#include <string.h>

#include <iostream>

#include <algorithm>

#include <vector>

#include <queue>

#include <set>

#include <map>

#include <string>

#include <math.h>

#include <stdlib.h>

#include <time.h>

#include <assert.h>

using namespace std;

const int maxnode = 550005;

const int MaxM = 999;

const int MaxN = 555;

int K,n,m;

void print(int x)

{

    printf("(%d %d)\n",x%(n+1)==0?

n:(x%(n+1)-1),x%(n+1)==0?

x/(n+1)-1:x/(n+1));

}

struct DLX

{

    int n,m,size;

    int U[maxnode],D[maxnode],R[maxnode],L[maxnode],Row[maxnode],Col[maxnode];

    int H[MaxN],S[MaxM];

    int ands,ans[MaxN],ANS;

    void init(int _n,int _m)

    {

        ANS=0x3f3f3f3f;

        n = _n;

        m = _m;

        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)

    {

        int i,j;

        L[R[c]]=L[c];

        R[L[c]]=R[c];

        for(i=D[c]; i!=c; i=D[i])

        {

            for(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)

    {

        int i,j;

        for(i=U[c]; i!=c; i=U[i])

        {

            for(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 v[maxnode];

    int f()

    {

        int ret = 0;

        for(int c = R[0]; c != 0; c = R[c])v[c] = true;

        for(int c = R[0]; c != 0; c = R[c])

            if(v[c])

            {

                ret++;

                v[c] = false;

                for(int i = D[c]; i != c; i = D[i])

                    for(int j = R[i]; j != i; j = R[j])

                        v[Col[j]] = false;

            }

        return ret;

    }

    void Dance(int d)

    {

        if(d+f()>=ANS) return;

        if(R[0] == 0)

        {

            ANS=min(ANS,d);

            return;

        }

        int c = R[0];

        for(int i = R[0]; i != 0; i = R[i])

            if(S[i] < S[c])

                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);

    }

} g;

int main()

{

    int cas,x1,y1,x2,y2,k;

    scanf("%d",&cas);

    while(cas--)

    {

        scanf("%d%d%d",&n,&m,&k);

        g.init(k,n*m);

        for(int q=1; q<=k; q++)

        {

            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++)

                    g.Link(q,j+(i-1)*m);

        }

        g.Dance(0);

        if(g.ANS==0x3f3f3f3f) g.ANS=-1;

        printf("%d\n",g.ANS);

    }

    return 0;

}