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，但由于矩阵拼接过程中，有公共的边，这里需要的技巧就是把矩阵的左边和下面截去一个单位。

#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;
}

ZOJ 3209 Treasure Map DLX